Displacement, Velocity, and Acceleration (CED Unit 1) Kinematics describes the motion of objects without considering the forces that cause motion.
Scalar vs Vector Quantities Quantity Type Description Displacement Vector Change in position, Δ r ⃗ \Delta \vec{r} Δ r Distance Scalar Total path length traveled Velocity Vector Rate of change of displacement Speed Scalar Rate of change of distance (magnitude of velocity) Acceleration Vector Rate of change of velocity
Definitions Using Calculus For position r ⃗ ( t ) \vec{r}(t) r ( t )
{ ⃗ v } ( t ) = { d { ⃗ r } } { d t } \vec\{v\}(t) = \frac\{d\vec\{r\}\}\{dt\} { v } ( t ) = d { { r }} { d t } { ⃗ a } ( t ) = { d { ⃗ v } } { d t } = { d 2 { ⃗ r } } { d t 2 } \vec\{a\}(t) = \frac\{d\vec\{v\}\}\{dt\} = \frac\{d^2\vec\{r\}\}\{dt^2\} { a } ( t ) = d { { v }} { d t } = d { 2 { r }} { d t 2 } Conversely, by integration:
{ ⃗ v } ( t ) = { ⃗ v } 0 + ∫ 0 t { ⃗ a } ( t ′ ) d t ′ \vec\{v\}(t) = \vec\{v\}_0 + \int_0^t \vec\{a\}(t')\, dt' { v } ( t ) = { v } 0 + ∫ 0 t { a } ( t ′ ) d t ′ { ⃗ r } ( t ) = { ⃗ r } 0 + ∫ 0 t { ⃗ v } ( t ′ ) d t ′ \vec\{r\}(t) = \vec\{r\}_0 + \int_0^t \vec\{v\}(t')\, dt' { r } ( t ) = { r } 0 + ∫ 0 t { v } ( t ′ ) d t ′ Why Calculus Is the Natural Language of Kinematics Velocity is the derivative of position. Acceleration is the derivative of velocity. These are not Empirical observations; they are the definitions of velocity and acceleration. The kinematic Equations are the results of integrating constant acceleration once (to get velocity) and Twice (to get position). Every result in kinematics follows from these definitions plus the Assumption of constant acceleration. If you understand the calculus, you understand the kinematics.
One-Dimensional Kinematics Constant Acceleration Equations When acceleration a a a
V ( t ) = v 0 + a t V(t) = v_0 + at V ( t ) = v 0 + a t X ( t ) = x 0 + v 0 t + { 1 } { 2 } a t 2 X(t) = x_0 + v_0 t + \frac\{1\}\{2\}at^2 X ( t ) = x 0 + v 0 t + 1 { } { 2 } a t 2 V 2 = v 0 2 + 2 a ( x − x 0 ) V^2 = v_0^2 + 2a(x - x_0) V 2 = v 0 2 + 2 a ( x − x 0 ) X − x 0 = { v 0 + v } { 2 } t X - x_0 = \frac\{v_0 + v\}\{2\} t X − x 0 = v { 0 + v } { 2 } t Derivation from Calculus Starting from a = d v d t = c o n s t a = \frac{dv}{dt} = \mathrm{const} a = d t d v = const
∫ { v 0 } { v } d v ′ = ∫ 0 t a d t ′ ⟹ v − v 0 = a t ⟹ v = v 0 + a t \int_\{v_0\}^\{v\} dv' = \int_0^t a\, dt' \implies v - v_0 = at \implies v = v_0 + at ∫ { v 0 } { v } d v ′ = ∫ 0 t a d t ′ ⟹ v − v 0 = a t ⟹ v = v 0 + a t Then v = d x d t v = \frac{dx}{dt} v = d t d x
∫ { x 0 } { x } d x ′ = ∫ 0 t ( v 0 + a t ′ ) d t ′ ⟹ x − x 0 = v 0 t + { 1 } { 2 } a t 2 \int_\{x_0\}^\{x\} dx' = \int_0^t (v_0 + at')\, dt' \implies x - x_0 = v_0 t + \frac\{1\}\{2\}at^2 ∫ { x 0 } { x } d x ′ = ∫ 0 t ( v 0 + a t ′ ) d t ′ ⟹ x − x 0 = v 0 t + 1 { } { 2 } a t 2 Eliminating Time: The v 2 v^2 v 2 To eliminate t t t t = ( v − v 0 ) / a t = (v - v_0)/a t = ( v − v 0 ) / a
X − x 0 = v 0 { v − v 0 } { a } + { 1 } { 2 } a ( { v − v 0 } { a } ) 2 X - x_0 = v_0 \frac\{v - v_0\}\{a\} + \frac\{1\}\{2\}a\left(\frac\{v - v_0\}\{a\}\right)^2 X − x 0 = v 0 v { − v 0 } { a } + 1 { } { 2 } a ( v { − v 0 } { a } ) 2 X − x 0 = { v 0 v − v 0 2 } { a } + { v 2 − 2 v 0 v + v 0 2 } { 2 a } X - x_0 = \frac\{v_0 v - v_0^2\}\{a\} + \frac\{v^2 - 2v_0 v + v_0^2\}\{2a\} X − x 0 = v { 0 v − v 0 2 } { a } + v { 2 − 2 v 0 v + v 0 2 } { 2 a } 2 a ( x − x 0 ) = 2 v 0 v − 2 v 0 2 + v 2 − 2 v 0 v + v 0 2 = v 2 − v 0 2 2a(x - x_0) = 2v_0 v - 2v_0^2 + v^2 - 2v_0 v + v_0^2 = v^2 - v_0^2 2 a ( x − x 0 ) = 2 v 0 v − 2 v 0 2 + v 2 − 2 v 0 v + v 0 2 = v 2 − v 0 2 V 2 = v 0 2 + 2 a ( x − x 0 ) V^2 = v_0^2 + 2a(x - x_0) V 2 = v 0 2 + 2 a ( x − x 0 ) This equation is useful because it relates velocity and displacement without requiring the time.
A car accelerates uniformly from rest at 3.0 m / s 2 3.0 \mathrm{ m/s^2} 3.0 m/ s 2
X = 0 + 0 ⋅ 5.0 + { 1 } { 2 } ( 3.0 ) ( 5.0 ) 2 = { 1 } { 2 } ( 3.0 ) ( 25 ) = 37.5 { m X = 0 + 0 \cdot 5.0 + \frac\{1\}\{2\}(3.0)(5.0)^2 = \frac\{1\}\{2\}(3.0)(25) = 37.5 \mathrm\{ m X = 0 + 0 ⋅ 5.0 + 1 { } { 2 } ( 3.0 ) ( 5.0 ) 2 = 1 { } { 2 } ( 3.0 ) ( 25 ) = 37.5 { m When to Use Each Equation Each of the four kinematic equations relates a different subset of the five variables (v v v v 0 v_0 v 0 a a a t t t x − x 0 x - x_0 x − x 0
Known variables Use equation v 0 v_0 v 0 a a a t t t v = v 0 + a t v = v_0 + at v = v 0 + a t v 0 v_0 v 0 a a a t t t x − x 0 = v 0 t + 1 2 a t 2 x - x_0 = v_0 t + \frac{1}{2}at^2 x − x 0 = v 0 t + 2 1 a t 2 v 0 v_0 v 0 a a a Δ x \Delta x Δ x v 2 = v 0 2 + 2 a Δ x v^2 = v_0^2 + 2a\Delta x v 2 = v 0 2 + 2 a Δ x v 0 v_0 v 0 v v v t t t Δ x = v 0 + v 2 t \Delta x = \frac{v_0 + v}{2} t Δ x = 2 v 0 + v t
The variable that does not appear in each equation tells you when to use it.
Free Fall Near Earth’s surface, the acceleration due to gravity is approximately:
G ≈ 9.8 { m / s 2 ≈ 32 { f t / s 2 G \approx 9.8 \mathrm\{ m/s^2 \approx 32 \mathrm\{ ft/s^2 G ≈ 9.8 { m / s 2 ≈ 32 { f t / s 2 Directed downward. In the standard coordinate system (upward positive), a = − g a = -g a = − g
Equations of Free Fall V = v 0 − g t V = v_0 - gt V = v 0 − g t Y = y 0 + v 0 t − { 1 } { 2 } g t 2 Y = y_0 + v_0 t - \frac\{1\}\{2\}gt^2 Y = y 0 + v 0 t − 1 { } { 2 } g t 2 V 2 = v 0 2 − 2 g ( y − y 0 ) V^2 = v_0^2 - 2g(y - y_0) V 2 = v 0 2 − 2 g ( y − y 0 ) Symmetry of Free Fall An object thrown upward and an object dropped from the same height take the same time to reach the Ground (if air resistance is negligible). This is because the motion is symmetric about the peak: The time to rise equals the time to fall, and the speed at any height on the way up equals the speed at the same height on the way down. The symmetry is a direct consequence of the equations being Quadratic in t t t
A ball is thrown upward from ground level with initial velocity 20 m / s 20 \mathrm{ m/s} 20 m/s
Maximum height: At the peak, v = 0 v = 0 v = 0
V 2 = v 0 2 − 2 g ( y − y 0 ) ⟹ 0 = ( 20 ) 2 − 2 ( 9.8 ) y ⟹ y = { 4 00 } { 19.6 } ≈ 20.4 { m V^2 = v_0^2 - 2g(y - y_0) \implies 0 = (20)^2 - 2(9.8)y \implies y = \frac\{400\}\{19.6\} \approx 20.4 \mathrm\{ m V 2 = v 0 2 − 2 g ( y − y 0 ) ⟹ 0 = ( 20 ) 2 − 2 ( 9.8 ) y ⟹ y = 4 { 00 } { 19.6 } ≈ 20.4 { m Time in air: The ball returns to y = 0 y = 0 y = 0
0 = 20 t − { 1 } { 2 } ( 9.8 ) t 2 ⟹ t ( 20 − 4.9 t ) = 0 ⟹ t = { 2 0 } { 4.9 } ≈ 4.08 { s 0 = 20t - \frac\{1\}\{2\}(9.8)t^2 \implies t(20 - 4.9t) = 0 \implies t = \frac\{20\}\{4.9\} \approx 4.08 \mathrm\{ s 0 = 20 t − 1 { } { 2 } ( 9.8 ) t 2 ⟹ t ( 20 − 4.9 t ) = 0 ⟹ t = 2 { 0 } { 4.9 } ≈ 4.08 { s Two-Dimensional Kinematics (Projectile Motion) (CED Unit 1) In projectile motion, the horizontal and vertical motions are independent.
Why Horizontal and Vertical Motions Are Independent The horizontal component of velocity has no effect on the vertical acceleration (which is always g g g
Horizontal Motion (constant velocity, no acceleration) X ( t ) = x 0 + v { 0 x } t = x 0 + ( v 0 cos θ ) t X(t) = x_0 + v_\{0x\} t = x_0 + (v_0 \cos\theta) t X ( t ) = x 0 + v { 0 x } t = x 0 + ( v 0 cos θ ) t V x ( t ) = v 0 cos θ { ( c o n s t a n t ) V_x(t) = v_0 \cos\theta \quad \mathrm\{(constant) V x ( t ) = v 0 cos θ {( co n s t an t ) Vertical Motion (free fall) Y ( t ) = y 0 + v { 0 y } t − { 1 } { 2 } g t 2 = y 0 + ( v 0 sin θ ) t − { 1 } { 2 } g t 2 Y(t) = y_0 + v_\{0y\} t - \frac\{1\}\{2\}gt^2 = y_0 + (v_0 \sin\theta) t - \frac\{1\}\{2\}gt^2 Y ( t ) = y 0 + v { 0 y } t − 1 { } { 2 } g t 2 = y 0 + ( v 0 sin θ ) t − 1 { } { 2 } g t 2 V y ( t ) = v 0 sin θ − g t V_y(t) = v_0 \sin\theta - gt V y ( t ) = v 0 sin θ − g t Key Results Time to reach maximum height: t u p = v 0 sin θ g \displaystyle t_{\mathrm{up}} = \frac{v_0 \sin\theta}{g} t up = g v 0 sin θ Maximum height: H = ( v 0 sin θ ) 2 2 g \displaystyle H = \frac{(v_0 \sin\theta)^2}{2g} H = 2 g ( v 0 sin θ ) 2 Total time of flight: T = 2 v 0 sin θ g \displaystyle T = \frac{2v_0 \sin\theta}{g} T = g 2 v 0 sin θ Range: R = v 0 2 sin 2 θ g \displaystyle R = \frac{v_0^2 \sin 2\theta}{g} R = g v 0 2 sin 2 θ Maximum range (for fixed v 0 v_0 v 0 θ = 45 ∘ \theta = 45^\circ θ = 4 5 ∘ The projectile lands when y = 0 y = 0 y = 0
0 = ( v 0 sin θ ) T − { 1 } { 2 } g T 2 ⟹ T = { 2 v 0 sin θ } { g } 0 = (v_0 \sin\theta) T - \frac\{1\}\{2\}gT^2 \implies T = \frac\{2v_0 \sin\theta\}\{g\} 0 = ( v 0 sin θ ) T − 1 { } { 2 } g T 2 ⟹ T = 2 { v 0 sin θ } { g } The horizontal distance at time T T T
R = ( v 0 cos θ ) T = ( v 0 cos θ ) ⋅ { 2 v 0 sin θ } { g } = { v 0 2 ⋅ 2 sin θ cos θ } { g } = { v 0 2 sin 2 θ } { g } R = (v_0 \cos\theta) T = (v_0 \cos\theta) \cdot \frac\{2v_0 \sin\theta\}\{g\} = \frac\{v_0^2 \cdot 2\sin\theta \cos\theta\}\{g\} = \frac\{v_0^2 \sin 2\theta\}\{g\} R = ( v 0 cos θ ) T = ( v 0 cos θ ) ⋅ 2 { v 0 sin θ } { g } = v { 0 2 ⋅ 2 sin θ cos θ } { g } = v { 0 2 sin 2 θ } { g } Using the double-angle identity sin 2 θ = 2 sin θ cos θ \sin 2\theta = 2\sin\theta \cos\theta sin 2 θ = 2 sin θ cos θ
Why Maximum Range Is at 45 Degrees The range R = ( v 0 2 / g ) sin 2 θ R = (v_0^2/g)\sin 2\theta R = ( v 0 2 / g ) sin 2 θ sin 2 θ \sin 2\theta sin 2 θ 2 θ = 90 ∘ 2\theta = 90^{\circ} 2 θ = 9 0 ∘ θ = 45 ∘ \theta = 45^{\circ} θ = 4 5 ∘ sin 2 θ = sin 2 ( 90 ∘ − θ ) \sin 2\theta = \sin 2(90^{\circ} - \theta) sin 2 θ = sin 2 ( 9 0 ∘ − θ ) 30 ∘ 30^{\circ} 3 0 ∘ 60 ∘ 60^{\circ} 6 0 ∘ 60 ∘ 60^{\circ} 6 0 ∘
The Trajectory Is a Parabola Eliminating t t t
T = { x } { v 0 cos θ } T = \frac\{x\}\{v_0 \cos\theta\} T = x { } { v 0 cos θ } Y = x tan θ − { g x 2 } { 2 v 0 2 cos 2 θ } Y = x\tan\theta - \frac\{gx^2\}\{2v_0^2\cos^2\theta\} Y = x tan θ − g { x 2 } { 2 v 0 2 cos 2 θ } This is a quadratic in x x x x 2 x^2 x 2
A projectile is launched at 50 m / s 50 \mathrm{ m/s} 50 m/s 37 ∘ 37^\circ 3 7 ∘
cos 37 ∘ ≈ 0.8 \cos 37^\circ \approx 0.8 cos 3 7 ∘ ≈ 0.8 sin 37 ∘ ≈ 0.6 \sin 37^\circ \approx 0.6 sin 3 7 ∘ ≈ 0.6
Range:
R = { ( 50 ) 2 sin 74 ∘ } { 9.8 } = { 2 500 × 0.9613 } { 9.8 } ≈ 245.2 { m R = \frac\{(50)^2 \sin 74^\circ\}\{9.8\} = \frac\{2500 \times 0.9613\}\{9.8\} \approx 245.2 \mathrm\{ m R = ( { 50 ) 2 sin 7 4 ∘ } { 9.8 } = 2 { 500 × 0.9613 } { 9.8 } ≈ 245.2 { m Maximum height:
H = { ( 50 × 0.6 ) 2 } { 2 × 9.8 } = { 9 00 } { 19.6 } ≈ 45.9 { m H = \frac\{(50 \times 0.6)^2\}\{2 \times 9.8\} = \frac\{900\}\{19.6\} \approx 45.9 \mathrm\{ m H = ( { 50 × 0.6 ) 2 } { 2 × 9.8 } = 9 { 00 } { 19.6 } ≈ 45.9 { m Projectile from a Height When a projectile is launched from a height h h h 2 v 0 sin θ / g 2v_0\sin\theta/g 2 v 0 sin θ / g t t t y ( t ) = 0 y(t) = 0 y ( t ) = 0 R = v 0 cos θ ⋅ t R = v_0\cos\theta \cdot t R = v 0 cos θ ⋅ t
Position, Velocity, and Acceleration as Graphs (CED Unit 1) From Graphs The slope of a position-time graph gives velocity. The slope of a velocity-time graph gives acceleration. The area under a velocity-time graph gives displacement. The area under an acceleration-time graph gives the change in velocity. Qualitative Analysis Positive slope on x x x t t t moving in the positive direction.Zero slope on x x x t t t at rest.Negative slope on x x x t t t moving in the negative direction.Curvature on x x x t t t indicates acceleration (concave up = positive a a a a a a Inflection Points and Velocity Extrema On an x x x t t t
An object moving in a circle of radius r r r v v v
Centripetal acceleration: a c = v 2 r \displaystyle a_c = \frac{v^2}{r} a c = r v 2 Period: T = 2 π r v \displaystyle T = \frac{2\pi r}{v} T = v 2 π r Frequency: f = 1 T = v 2 π r \displaystyle f = \frac{1}{T} = \frac{v}{2\pi r} f = T 1 = 2 π r v Angular Quantities ω = { d θ } { d t } = { v } { r } \omega = \frac\{d\theta\}\{dt\} = \frac\{v\}\{r\} ω = d { θ } { d t } = v { } { r } A c = ω 2 r = { v 2 } { r } A_c = \omega^2 r = \frac\{v^2\}\{r\} A c = ω 2 r = v { 2 } { r } Derivation of Centripetal Acceleration Consider a particle moving at constant speed v v v r r r Δ t \Delta t Δ t Δ θ \Delta\theta Δ θ v v v
The change in velocity Δ v ⃗ \Delta\vec{v} Δ v v ⃗ \vec{v} v
∣ Δ { ⃗ v } ∣ = v ⋅ Δ θ |\Delta\vec\{v\}| = v \cdot \Delta\theta ∣Δ { v } ∣ = v ⋅ Δ θ Since Δ θ = v Δ t r \Delta\theta = \frac{v \Delta t}{r} Δ θ = r v Δ t
∣ Δ { ⃗ v } ∣ = v ⋅ { v Δ t } { r } = { v 2 Δ t } { r } |\Delta\vec\{v\}| = v \cdot \frac\{v \Delta t\}\{r\} = \frac\{v^2 \Delta t\}\{r\} ∣Δ { v } ∣ = v ⋅ v { Δ t } { r } = v { 2 Δ t } { r } The centripetal acceleration is:
A c = lim { Δ t → 0 } { ∣ Δ { ⃗ v } ∣ } { Δ t } = { v 2 } { r } A_c = \lim_\{\Delta t \to 0\} \frac\{|\Delta\vec\{v\}|\}\{\Delta t\} = \frac\{v^2\}\{r\} A c = { lim Δ t → 0 } ∣ { Δ { v } ∣ } { Δ t } = v { 2 } { r } Why Centripetal Acceleration Does Not Change Speed Centripetal acceleration is always perpendicular to the velocity. Since the dot product a ⃗ c ⋅ v ⃗ = 0 \vec{a}_c \cdot \vec{v} = 0 a c ⋅ v = 0 P = F ⃗ ⋅ v ⃗ = 0 P = \vec{F} \cdot \vec{v} = 0 P = F ⋅ v = 0 1 2 m v 2 \frac{1}{2}mv^2 2 1 m v 2
A car rounds a curve of radius 50 m at 15 m / s 15 \mathrm{ m/s} 15 m/s
A c = { v 2 } { r } = { ( 15 ) 2 } { 50 } = { 2 25 } { 50 } = 4.5 { m / s 2 A_c = \frac\{v^2\}\{r\} = \frac\{(15)^2\}\{50\} = \frac\{225\}\{50\} = 4.5 \mathrm\{ m/s^2 A c = v { 2 } { r } = ( { 15 ) 2 } { 50 } = 2 { 25 } { 50 } = 4.5 { m / s 2 When acceleration is not constant, use calculus.
Velocity from Variable Acceleration V ( t ) = v 0 + ∫ 0 t a ( t ′ ) d t ′ V(t) = v_0 + \int_0^t a(t')\, dt' V ( t ) = v 0 + ∫ 0 t a ( t ′ ) d t ′ Position from Variable Velocity X ( t ) = x 0 + ∫ 0 t v ( t ′ ) d t ′ X(t) = x_0 + \int_0^t v(t')\, dt' X ( t ) = x 0 + ∫ 0 t v ( t ′ ) d t ′ Displacement as Area Under the Curve When a ( t ) a(t) a ( t ) v ( t ) v(t) v ( t ) v v v t t t
The velocity of a particle is v ( t ) = 3 t 2 − 6 t + 2 m / s v(t) = 3t^2 - 6t + 2 \mathrm{ m/s} v ( t ) = 3 t 2 − 6 t + 2 m/s t = 0 t = 0 t = 0 t = 3 t = 3 t = 3 t = 2 t = 2 t = 2
Displacement:
Δ x = ∫ 0 3 v ( t ) d t = ∫ 0 3 ( 3 t 2 − 6 t + 2 ) d t = [ t 3 − 3 t 2 + 2 t ] 0 3 = 27 − 27 + 6 = 6 { m \Delta x = \int_0^3 v(t)\, dt = \int_0^3 (3t^2 - 6t + 2)\, dt = \left[t^3 - 3t^2 + 2t\right]_0^3 = 27 - 27 + 6 = 6 \mathrm\{ m Δ x = ∫ 0 3 v ( t ) d t = ∫ 0 3 ( 3 t 2 − 6 t + 2 ) d t = [ t 3 − 3 t 2 + 2 t ] 0 3 = 27 − 27 + 6 = 6 { m Acceleration:
A ( t ) = { d v } { d t } = 6 t − 6 ⟹ a ( 2 ) = 6 ( 2 ) − 6 = 6 { m / s 2 A(t) = \frac\{dv\}\{dt\} = 6t - 6 \implies a(2) = 6(2) - 6 = 6 \mathrm\{ m/s^2 A ( t ) = d { v } { d t } = 6 t − 6 ⟹ a ( 2 ) = 6 ( 2 ) − 6 = 6 { m / s 2 Worked Example: Reversing the Problem A particle has acceleration a ( t ) = 12 t a(t) = 12t a ( t ) = 12 t 2 ^2 2 v ( 0 ) = 5 v(0) = 5 v ( 0 ) = 5 x ( 0 ) = 0 x(0) = 0 x ( 0 ) = 0 v ( t ) v(t) v ( t ) x ( t ) x(t) x ( t )
V ( t ) = v ( 0 ) + ∫ 0 t 12 t ′ d t ′ = 5 + 6 t 2 V(t) = v(0) + \int_0^t 12t'\, dt' = 5 + 6t^2 V ( t ) = v ( 0 ) + ∫ 0 t 12 t ′ d t ′ = 5 + 6 t 2 X ( t ) = x ( 0 ) + ∫ 0 t ( 5 + 6 t ′ 2 ) d t ′ = 5 t + 2 t 3 X(t) = x(0) + \int_0^t (5 + 6t'^2)\, dt' = 5t + 2t^3 X ( t ) = x ( 0 ) + ∫ 0 t ( 5 + 6 t ′2 ) d t ′ = 5 t + 2 t 3 Relative Motion In one dimension with two reference frames S S S S ′ S' S ′ S ′ S' S ′ v S ′ / S v_{S'/S} v S ′ / S S S S
{ ⃗ v } { P / S } = { ⃗ v } { P / S ′ } + { ⃗ v } { S ′ / S } \vec\{v\}_\{P/S\} = \vec\{v\}_\{P/S'\} + \vec\{v\}_\{S'/S\} { v } { P / S } = { v } { P / S ′ } + { v } { S ′ / S } This is the Galilean velocity addition formula.
Relative Motion in Two Dimensions In two dimensions, the velocity addition is done component by component:
V { P / x } = v { P / S ′ } cos α + v { S ′ / S } cos β V_\{P/x\} = v_\{P/S'\}\cos\alpha + v_\{S'/S\}\cos\beta V { P / x } = v { P / S ′ } cos α + v { S ′ / S } cos β V { P / y } = v { P / S ′ } sin α + v { S ′ / S } sin β V_\{P/y\} = v_\{P/S'\}\sin\alpha + v_\{S'/S\}\sin\beta V { P / y } = v { P / S ′ } sin α + v { S ′ / S } sin β Where α \alpha α S ′ S' S ′ β \beta β S ′ S' S ′ S S S
A boat heads directly across a river 100 m wide at 4 m / s 4 \mathrm{ m/s} 4 m/s 3 m / s 3 \mathrm{ m/s} 3 m/s
The boat’s velocity relative to the shore has two components:
V x = 3 { m / s { ( d o w n s t r e a m ) , v y = 4 { m / s { ( a c r o s s ) V_x = 3 \mathrm\{ m/s \quad \mathrm\{(downstream), \qquad v_y = 4 \mathrm\{ m/s \quad \mathrm\{(across) V x = 3 { m / s {( d o w n s t r e am ) , v y = 4 { m / s {( a cr oss ) Time to cross: t = 100 4 = 25 s \displaystyle t = \frac{100}{4} = 25 \mathrm{ s} t = 4 100 = 25 s
Downstream distance: d = 3 × 25 = 75 m d = 3 \times 25 = 75 \mathrm{ m} d = 3 × 25 = 75 m
Crossing the River Perpendicular to the Bank To land directly across the river, the boat must angle upstream so that the downstream component of its velocity relative to the water cancels the river current. If the boat has speed v b v_b v b v r v_r v r
sin θ = { v r } { v b } \sin\theta = \frac\{v_r\}\{v_b\} sin θ = v { r } { v b } Where θ \theta θ v b > v r v_b \gt v_r v b > v r
Intuition Kinematics is the language of motion , it describes what is happening without asking why . Think of it as the GPS of physics: it tells you position, velocity, and acceleration, but not the forces causing them.
The calculus connection: Velocity is the derivative of position, acceleration is the derivative of velocity. This isn’t just a formula, it’s the definition . The kinematic equations for constant acceleration are directly what you get when you integrate a = const a = \text{const} a = const v v v x x x
Projectile motion intuition: The horizontal and vertical motions are completely independent. Gravity only acts vertically, so the horizontal velocity never changes. This means you can analyse each direction separately using the same equations, a powerful simplification.
Circular motion intuition: Even at constant speed, an object moving in a circle is accelerating, because acceleration is the rate of change of the velocity vector , and the direction keeps changing. This centripetal acceleration always points toward the center, like a tether pulling the object inward.
Common Pitfalls Confusing displacement with distance. Displacement is a vector (net change in position); distance is a scalar (total path length).Ignoring the sign convention. In free fall, choosing “up as positive” means a = − g a = -g a = − g Treating horizontal and vertical components as dependent. In projectile motion, the horizontal motion has constant velocity (no horizontal acceleration, ignoring air resistance).Using the constant acceleration equations when acceleration is not constant. Use calculus (integration) instead.Confusing speed and velocity. Speed is the magnitude of velocity. An object can have constant speed but changing velocity (e.g., uniform circular motion).Forgetting that centripetal acceleration changes direction, not speed. In uniform circular motion, the speed is constant but the velocity vector changes direction continuously.Incorrectly applying the range formula when the launch and landing heights differ.Forgetting that the kinematic equations assume constant acceleration. If the problem involves variable acceleration (a spring, a pendulum, a rocket losing mass), you must use calculus.Mixing up initial and final values. v 0 v_0 v 0 v v v Assuming the maximum height occurs at half the total time. This is only true when the launch and landing heights are the same.Practice Questions A ball is dropped from a height of 80 m. How long does it take to reach the ground, and what is its speed at impact?
A projectile is launched from a cliff 60 m high at 30 m / s 30 \mathrm{ m/s} 30 m/s 40 ∘ 40^\circ 4 0 ∘
A particle has acceleration a ( t ) = 6 t m / s 2 a(t) = 6t \mathrm{ m/s^2} a ( t ) = 6 t m/ s 2 v ( 0 ) = − 4 m / s v(0) = -4 \mathrm{ m/s} v ( 0 ) = − 4 m/s x ( 0 ) = 2 m x(0) = 2 \mathrm{ m} x ( 0 ) = 2 m v ( t ) v(t) v ( t ) x ( t ) x(t) x ( t ) t = 0 t = 0 t = 0 t = 3 t = 3 t = 3
A car accelerates from rest at 2.0 m / s 2 2.0 \mathrm{ m/s^2} 2.0 m/ s 2
An object moves along the x x x v ( t ) = t 3 − 6 t 2 + 9 t v(t) = t^3 - 6t^2 + 9t v ( t ) = t 3 − 6 t 2 + 9 t t = 0 t = 0 t = 0 t = 5 t = 5 t = 5
A satellite orbits Earth at an altitude of 300 km in a circular orbit. If the orbital period is 90.5 minutes, find the orbital speed and centripetal acceleration. (Earth’s radius = 6371 6371 6371
A swimmer can swim at 2.0 m / s 2.0 \mathrm{ m/s} 2.0 m/s 1.5 m / s 1.5 \mathrm{ m/s} 1.5 m/s
Derive the formula for the maximum height of a projectile using calculus.
A ball is thrown from the top of a 50 m building at 20 m / s 20 \mathrm{ m/s} 20 m/s 30 ∘ 30^\circ 3 0 ∘
The position of a particle is x ( t ) = t 3 − 6 t 2 + 9 t + 1 x(t) = t^3 - 6t^2 + 9t + 1 x ( t ) = t 3 − 6 t 2 + 9 t + 1 t = 0 t = 0 t = 0 t = 5 t = 5 t = 5
Motion Graphs: Advanced Analysis Determining Velocity from a Displacement-Time Graph For a curved x x x t t t
Worked Example: Interpreting a Velocity-Time Graph A particle moves along the x x x v ( t ) = 4 t − t 2 v(t) = 4t - t^2 v ( t ) = 4 t − t 2 0 ≤ t ≤ 5 0 \le t \le 5 0 ≤ t ≤ 5
Velocity at t = 2 t = 2 t = 2 v ( 2 ) = 8 − 4 = 4 v(2) = 8 - 4 = 4 v ( 2 ) = 8 − 4 = 4
Acceleration: a ( t ) = d v d t = 4 − 2 t a(t) = \frac{dv}{dt} = 4 - 2t a ( t ) = d t d v = 4 − 2 t t = 2 t = 2 t = 2 a = 0 a = 0 a = 0
Displacement from t = 0 t = 0 t = 0 t = 4 t = 4 t = 4
Δ x = ∫ 0 4 ( 4 t − t 2 ) d t = [ 2 t 2 − t 3 3 ] 0 4 = 32 − 64 3 = 32 − 21.33 = 10.67 m \Delta x = \int_0^4 (4t - t^2)\, dt = \left[2t^2 - \frac{t^3}{3}\right]_0^4 = 32 - \frac{64}{3} = 32 - 21.33 = 10.67 \mathrm{ m} Δ x = ∫ 0 4 ( 4 t − t 2 ) d t = [ 2 t 2 − 3 t 3 ] 0 4 = 32 − 3 64 = 32 − 21.33 = 10.67 m Total distance from t = 0 t = 0 t = 0 t = 5 t = 5 t = 5 The velocity is zero at t = 0 t = 0 t = 0 t = 4 t = 4 t = 4 0 \lt t \lt 4$$v \gt 0 . For 4 \lt t \lt 5$$v \lt 0 .
d = ∫ 0 4 ( 4 t − t 2 ) d t + ∣ ∫ 4 5 ( 4 t − t 2 ) d t ∣ = 10.67 + ∣ [ 2 t 2 − t 3 3 ] 4 5 ∣ d = \int_0^4 (4t - t^2)\, dt + \left|\int_4^5 (4t - t^2)\, dt\right| = 10.67 + \left|\left[2t^2 - \frac{t^3}{3}\right]_4^5\right| d = ∫ 0 4 ( 4 t − t 2 ) d t + ∫ 4 5 ( 4 t − t 2 ) d t = 10.67 + [ 2 t 2 − 3 t 3 ] 4 5 = 10.67 + ∣ ( 50 − 41.67 ) − ( 32 − 21.33 ) ∣ = 10.67 + ∣ 8.33 − 10.67 ∣ = 10.67 + 2.33 = 13.0 m = 10.67 + \left|(50 - 41.67) - (32 - 21.33)\right| = 10.67 + |8.33 - 10.67| = 10.67 + 2.33 = 13.0 \mathrm{ m} = 10.67 + ∣ ( 50 − 41.67 ) − ( 32 − 21.33 ) ∣ = 10.67 + ∣8.33 − 10.67∣ = 10.67 + 2.33 = 13.0 m Period, Frequency, and Angular Velocity Relationships For an object in uniform circular motion:
ω = 2 π f = 2 π T , v = 2 π r T = 2 π r f , a c = 4 π 2 r T 2 = ω 2 r = v 2 r \omega = 2\pi f = \frac{2\pi}{T}, \qquad v = \frac{2\pi r}{T} = 2\pi rf, \qquad a_c = \frac{4\pi^2 r}{T^2} = \omega^2 r = \frac{v^2}{r} ω = 2 π f = T 2 π , v = T 2 π r = 2 π r f , a c = T 2 4 π 2 r = ω 2 r = r v 2 Worked Example: Geostationary Orbit A geostationary satellite orbits above the equator with a period of 24 hours. Given Earth’s radius R E = 6.371 × 10 6 R_E = 6.371 \times 10^6 R E = 6.371 × 1 0 6 M E = 5.97 × 10 24 M_E = 5.97 \times 10^{24} M E = 5.97 × 1 0 24
T = 86400 s T = 86400 \mathrm{ s} T = 86400 s T 2 = 4 π 2 r 3 G M T^2 = \frac{4\pi^2 r^3}{GM} T 2 = GM 4 π 2 r 3 r 3 = G M T 2 4 π 2 = 6.674 × 10 − 11 × 5.97 × 10 24 × 86400 2 4 π 2 r^3 = \frac{GMT^2}{4\pi^2} = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 86400^2}{4\pi^2} r 3 = 4 π 2 GM T 2 = 4 π 2 6.674 × 1 0 − 11 × 5.97 × 1 0 24 × 8640 0 2 r 3 = 2.987 × 10 24 39.48 = 7.566 × 10 22 r^3 = \frac{2.987 \times 10^{24}}{39.48} = 7.566 \times 10^{22} r 3 = 39.48 2.987 × 1 0 24 = 7.566 × 1 0 22 r = 7.566 × 10 22 3 = 4.225 × 10 7 m r = \sqrt[3]{7.566 \times 10^{22}} = 4.225 \times 10^7 \mathrm{ m} r = 3 7.566 × 1 0 22 = 4.225 × 1 0 7 m h = r − R E = 4.225 × 10 7 − 6.371 × 10 6 = 3.588 × 10 7 m ≈ 35880 k m h = r - R_E = 4.225 \times 10^7 - 6.371 \times 10^6 = 3.588 \times 10^7 \mathrm{ m \approx 35880 \mathrm{ km}} h = r − R E = 4.225 × 1 0 7 − 6.371 × 1 0 6 = 3.588 × 1 0 7 m ≈ 35880 km This is approximately 36000 km above Earth’s surface, the standard geostationary orbit altitude.
Worked Example: Position, Velocity, and Acceleration from a Single Function The position of a particle is x ( t ) = t 4 − 4 t 3 + 6 t 2 x(t) = t^4 - 4t^3 + 6t^2 x ( t ) = t 4 − 4 t 3 + 6 t 2
v ( t ) = d x d t = 4 t 3 − 12 t 2 + 12 t v(t) = \frac{dx}{dt} = 4t^3 - 12t^2 + 12t v ( t ) = d t d x = 4 t 3 − 12 t 2 + 12 t a ( t ) = d v d t = 12 t 2 − 24 t + 12 = 12 ( t 2 − 2 t + 1 ) = 12 ( t − 1 ) 2 a(t) = \frac{dv}{dt} = 12t^2 - 24t + 12 = 12(t^2 - 2t + 1) = 12(t-1)^2 a ( t ) = d t d v = 12 t 2 − 24 t + 12 = 12 ( t 2 − 2 t + 1 ) = 12 ( t − 1 ) 2 The acceleration is always non-negative (it is zero at t = 1 t = 1 t = 1
Times when at rest: v ( t ) = 0 ⟹ 4 t ( t 2 − 3 t + 3 ) = 0 v(t) = 0 \implies 4t(t^2 - 3t + 3) = 0 v ( t ) = 0 ⟹ 4 t ( t 2 − 3 t + 3 ) = 0 t 2 − 3 t + 3 = 0 t^2 - 3t + 3 = 0 t 2 − 3 t + 3 = 0 9 − 12 = − 3 < 0 9 - 12 = -3 \lt 0 9 − 12 = − 3 < 0 t = 0 t = 0 t = 0
The particle is at rest only at t = 0 t = 0 t = 0
Worked Example: Piecewise-Defined Acceleration A particle starts from rest. For 0 ≤ t ≤ 3 0 \le t \le 3 0 ≤ t ≤ 3 a = 6 t a = 6t a = 6 t 2 ^2 2 t > 3 t \gt 3 t > 3 a = 0 a = 0 a = 0
Phase 1 (0 ≤ t ≤ 3 0 \le t \le 3 0 ≤ t ≤ 3
v ( t ) = ∫ 0 t 6 t ′ d t ′ = 3 t 2 v(t) = \int_0^t 6t'\, dt' = 3t^2 v ( t ) = ∫ 0 t 6 t ′ d t ′ = 3 t 2 x ( t ) = ∫ 0 t 3 t ′ 2 d t ′ = t 3 x(t) = \int_0^t 3t'^2\, dt' = t^3 x ( t ) = ∫ 0 t 3 t ′2 d t ′ = t 3 At t = 3 t = 3 t = 3 v = 27 v = 27 v = 27 x = 27 x = 27 x = 27
Phase 2 (t > 3 t \gt 3 t > 3 Constant velocity v = 27 v = 27 v = 27
x ( t ) = 27 + 27 ( t − 3 ) x(t) = 27 + 27(t - 3) x ( t ) = 27 + 27 ( t − 3 ) Summary Table: Kinematic Equations Equation Missing Variable Best Used When v = v 0 + a t v = v_0 + at v = v 0 + a t x − x 0 x - x_0 x − x 0 You know initial velocity, acceleration, time x − x 0 = v 0 t + 1 2 a t 2 x - x_0 = v_0 t + \frac{1}{2}at^2 x − x 0 = v 0 t + 2 1 a t 2 v v v You know initial velocity, acceleration, time v 2 = v 0 2 + 2 a ( x − x 0 ) v^2 = v_0^2 + 2a(x - x_0) v 2 = v 0 2 + 2 a ( x − x 0 ) t t t You know velocities and displacement x − x 0 = v 0 + v 2 t x - x_0 = \frac{v_0 + v}{2}t x − x 0 = 2 v 0 + v t a a a You know velocities and time (uniform acceleration)
Common Pitfalls: Extended Assuming that the magnitude of acceleration equals g g g The magnitude is g g g g g g g g g
Using the wrong reference frame in relative motion problems. Always identify the observer’s frame and the moving frame explicitly. Write the velocity addition formula carefully.
Forgetting that in 2D projectile motion, the time determines everything. Once you find the time of flight (from the vertical motion), you can find the horizontal range. You cannot find the range without first finding the time.
Confusing angular displacement with linear displacement in circular motion problems. Angular displacement θ \theta θ s = r θ s = r\theta s = r θ
Assuming that a = v 2 / r a = v^2/r a = v 2 / r The formula a c = v 2 / r a_c = v^2/r a c = v 2 / r a t = d v / d t a_t = dv/dt a t = d v / d t
Practice Questions (Additional) A particle moves along the x x x a ( t ) = 8 t − 6 a(t) = 8t - 6 a ( t ) = 8 t − 6 2 ^2 2 t = 0 t = 0 t = 0 x = 2 x = 2 x = 2 v = − 3 v = -3 v = − 3 t = 0 t = 0 t = 0 t = 3 t = 3 t = 3
A projectile is launched from ground level at 40 m / s 40 \mathrm{ m/s} 40 m/s 120 m 120 \mathrm{ m} 120 m
A car accelerates from rest at 3 m / s 2 3 \mathrm{ m/s^2} 3 m/ s 2 4 s 4 \mathrm{ s} 4 s 6 s 6 \mathrm{ s} 6 s 3 s 3 \mathrm{ s} 3 s
Two boats leave the same point simultaneously. Boat A heads north at 6 m / s 6 \mathrm{ m/s} 6 m/s 8 m / s 8 \mathrm{ m/s} 8 m/s
An astronaut on the Moon throws a ball vertically upward at 15 m / s 15 \mathrm{ m/s} 15 m/s g M o o n = 1.62 m / s 2 g_{\mathrm{Moon}} = 1.62 \mathrm{ m/s^2} g Moon = 1.62 m/ s 2
Extended Worked Examples Example 16: Relative Motion in Two Dimensions Airplane A flies north at 250 m / s 250 \mathrm{ m/s} 250 m/s 200 m / s 200 \mathrm{ m/s} 200 m/s 60 ∘ 60^\circ 6 0 ∘
Step 1: Write velocity vectors
v ⃗ A = 250 j ^ m / s \vec{v}_A = 250\hat{j} \mathrm{ m/s} v A = 250 j ^ m/s v ⃗ B = 200 sin 60 ° i ^ + 200 cos 60 ° j ^ = 173.2 i ^ + 100 j ^ m / s \vec{v}_B = 200\sin 60°\hat{i} + 200\cos 60°\hat{j} = 173.2\hat{i} + 100\hat{j} \mathrm{ m/s} v B = 200 sin 60° i ^ + 200 cos 60° j ^ = 173.2 i ^ + 100 j ^ m/s Step 2: Relative velocity
v ⃗ A B = v ⃗ A − v ⃗ B = − 173.2 i ^ + ( 250 − 100 ) j ^ = − 173.2 i ^ + 150 j ^ m / s \vec{v}_{AB} = \vec{v}_A - \vec{v}_B = -173.2\hat{i} + (250 - 100)\hat{j} = -173.2\hat{i} + 150\hat{j} \mathrm{ m/s} v A B = v A − v B = − 173.2 i ^ + ( 250 − 100 ) j ^ = − 173.2 i ^ + 150 j ^ m/s Step 3: Magnitude and direction
∣ v ⃗ A B ∣ = 173.2 2 + 150 2 = 29998 + 22500 = 52498 = 229.1 m / s |\vec{v}_{AB}| = \sqrt{173.2^2 + 150^2} = \sqrt{29998 + 22500} = \sqrt{52498} = 229.1 \mathrm{ m/s} ∣ v A B ∣ = 173. 2 2 + 15 0 2 = 29998 + 22500 = 52498 = 229.1 m/s θ = arctan ( 150 − 173.2 ) = 180 ° − 40.9 ° = 139.1 ° f r o m e a s t \theta = \arctan\left(\frac{150}{-173.2}\right) = 180° - 40.9° = 139.1° \mathrm{ from east} θ = arctan ( − 173.2 150 ) = 180° − 40.9° = 139.1° fromeast So the velocity of A relative to B is 229.1 m / s 229.1 \mathrm{ m/s} 229.1 m/s 139.1 ∘ 139.1^\circ 139. 1 ∘ 50.9 ∘ 50.9^\circ 50. 9 ∘
The velocity of a particle is given by v ( t ) = t 3 − 6 t 2 + 9 t v(t) = t^3 - 6t^2 + 9t v ( t ) = t 3 − 6 t 2 + 9 t 0 ≤ t ≤ 5 s 0 \le t \le 5 \mathrm{ s} 0 ≤ t ≤ 5 s
Find (a) when the particle is at rest, (b) the total distance travelled, and (c) the displacement.
Step 1: Find when the particle is at rest
v ( t ) = t 3 − 6 t 2 + 9 t = t ( t 2 − 6 t + 9 ) = t ( t − 3 ) 2 = 0 v(t) = t^3 - 6t^2 + 9t = t(t^2 - 6t + 9) = t(t - 3)^2 = 0 v ( t ) = t 3 − 6 t 2 + 9 t = t ( t 2 − 6 t + 9 ) = t ( t − 3 ) 2 = 0 t = 0 t = 0 t = 0 t = 3 s t = 3 \mathrm{ s} t = 3 s
Step 2: Determine the sign of v v v
0 < t < 3 0 \lt t \lt 3 0 < t < 3 t = 1 t = 1 t = 1 v = 1 − 6 + 9 = 4 > 0 v = 1 - 6 + 9 = 4 \gt 0 v = 1 − 6 + 9 = 4 > 0 t > 3 t \gt 3 t > 3 t = 4 t = 4 t = 4 v = 64 − 96 + 36 = 4 > 0 v = 64 - 96 + 36 = 4 \gt 0 v = 64 − 96 + 36 = 4 > 0 The particle never moves backward. The velocity is always non-negative (zero only at t = 0 t = 0 t = 0 t = 3 t = 3 t = 3
Step 3: Total distance = displacement (since v ≥ 0 v \ge 0 v ≥ 0
s ( 5 ) − s ( 0 ) = ∫ 0 5 ( t 3 − 6 t 2 + 9 t ) d t = [ t 4 4 − 2 t 3 + 9 t 2 2 ] 0 5 s(5) - s(0) = \int_0^5 (t^3 - 6t^2 + 9t) \, dt = \left[ \frac{t^4}{4} - 2t^3 + \frac{9t^2}{2} \right]_0^5 s ( 5 ) − s ( 0 ) = ∫ 0 5 ( t 3 − 6 t 2 + 9 t ) d t = [ 4 t 4 − 2 t 3 + 2 9 t 2 ] 0 5 = 625 4 − 250 + 225 2 = 156.25 − 250 + 112.5 = 18.75 m = \frac{625}{4} - 250 + \frac{225}{2} = 156.25 - 250 + 112.5 = 18.75 \mathrm{ m} = 4 625 − 250 + 2 225 = 156.25 − 250 + 112.5 = 18.75 m Step 4: Acceleration at key points
a ( t ) = 3 t 2 − 12 t + 9 = 3 ( t 2 − 4 t + 3 ) = 3 ( t − 1 ) ( t − 3 ) a(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t - 1)(t - 3) a ( t ) = 3 t 2 − 12 t + 9 = 3 ( t 2 − 4 t + 3 ) = 3 ( t − 1 ) ( t − 3 ) At t = 0 t = 0 t = 0 a = 9 m / s 2 a = 9 \mathrm{ m/s^2} a = 9 m/ s 2 t = 3 t = 3 t = 3 a = 3 ( 9 − 12 + 3 ) = 0 m / s 2 a = 3(9 - 12 + 3) = 0 \mathrm{ m/s^2} a = 3 ( 9 − 12 + 3 ) = 0 m/ s 2
Even though the particle is stationary at t = 3 s t = 3 \mathrm{ s} t = 3 s
There, so the particle does not reverse direction. Compare this to projectile motion at the top of the trajectory where v = 0 v = 0 v = 0 a = g ≠ 0 a = g \ne 0 a = g = 0
Example 18: Kinematics with Air Resistance A 70 k g 70 \mathrm{ kg} 70 kg F D = k v 2 F_D = kv^2 F D = k v 2 k = 0.25 k g / m k = 0.25 \mathrm{ kg/m} k = 0.25 kg/m 10 s 10 \mathrm{ s} 10 s 10 s 10 \mathrm{ s} 10 s
Step 1: Terminal velocity
At terminal velocity: m g = k v T 2 mg = kv_T^2 m g = k v T 2
v T = m g k = 70 × 9.8 0.25 = 2744 = 52.4 m / s v_T = \sqrt{\frac{mg}{k}} = \sqrt{\frac{70 \times 9.8}{0.25}} = \sqrt{2744} = 52.4 \mathrm{ m/s} v T = k m g = 0.25 70 × 9.8 = 2744 = 52.4 m/s Step 2: Analytical solution
The equation of motion is m d v d t = m g − k v 2 m\frac{dv}{dt} = mg - kv^2 m d t d v = m g − k v 2
This has the solution:
v ( t ) = v T tanh ( g t v T ) v(t) = v_T \tanh\left(\frac{gt}{v_T}\right) v ( t ) = v T tanh ( v T g t ) v ( 10 ) = 52.4 × tanh ( 9.8 × 10 52.4 ) = 52.4 × tanh ( 1.870 ) v(10) = 52.4 \times \tanh\left(\frac{9.8 \times 10}{52.4}\right) = 52.4 \times \tanh(1.870) v ( 10 ) = 52.4 × tanh ( 52.4 9.8 × 10 ) = 52.4 × tanh ( 1.870 ) tanh ( 1.870 ) = e 3.74 − e − 3.74 e 3.74 + e − 3.74 = 42.10 − 0.0238 42.10 + 0.0238 = 0.9989 \tanh(1.870) = \frac{e^{3.74} - e^{-3.74}}{e^{3.74} + e^{-3.74}} = \frac{42.10 - 0.0238}{42.10 + 0.0238} = 0.9989 tanh ( 1.870 ) = e 3.74 + e − 3.74 e 3.74 − e − 3.74 = 42.10 + 0.0238 42.10 − 0.0238 = 0.9989 v ( 10 ) = 52.4 × 0.9989 = 52.3 m / s v(10) = 52.4 \times 0.9989 = 52.3 \mathrm{ m/s} v ( 10 ) = 52.4 × 0.9989 = 52.3 m/s The skydiver has essentially reached terminal velocity after 10 s 10 \mathrm{ s} 10 s
Step 3: Distance fallen
y ( t ) = v T 2 g ln ( cosh ( g t v T ) ) y(t) = \frac{v_T^2}{g}\ln\left(\cosh\left(\frac{gt}{v_T}\right)\right) y ( t ) = g v T 2 ln ( cosh ( v T g t ) ) y ( 10 ) = 2744 9.8 ln ( cosh ( 1.870 ) ) = 280 × ln ( 3.263 ) = 280 × 1.183 = 331.2 m y(10) = \frac{2744}{9.8}\ln(\cosh(1.870)) = 280 \times \ln(3.263) = 280 \times 1.183 = 331.2 \mathrm{ m} y ( 10 ) = 9.8 2744 ln ( cosh ( 1.870 )) = 280 × ln ( 3.263 ) = 280 × 1.183 = 331.2 m Common Pitfalls Extended Pitfall 6: Using the Wrong Sign Convention in Free Fall Always define your positive direction before solving. If upward is positive, then a = − g a = -g a = − g
Pitfall 7: Confusing Average Velocity with Average Speed Average velocity = \frac{\mathrm{displacement}{\mathrm{time} (a vector, can be zero for round Trips). Average speed = \frac{\mathrm{total distance}{\mathrm{time} (a scalar, always positive). They are equal only for motion in one direction without reversing.
Pitfall 8: Assuming Graph Slopes Give Instantaneous Values Everywhere The slope of a displacement-time graph gives instantaneous velocity. The slope of a velocity-time Graph gives instantaneous acceleration. However, for curved graphs, the slope changes continuously. You must draw a tangent at the specific point to find the instantaneous value — the average slope between two points is not the same as the instantaneous slope.
Additional Practice Problems A ball is thrown upward from the top of a 50 m 50 \mathrm{ m} 50 m 20 m / s 20 \mathrm{ m/s} 20 m/s
A police car chasing a speeding motorist. The motorist passes at 30 m / s 30 \mathrm{ m/s} 30 m/s 2 s 2 \mathrm{ s} 2 s 3 m / s 2 3 \mathrm{ m/s^2} 3 m/ s 2
A projectile is launched from a cliff of height 80 m 80 \mathrm{ m} 80 m 50 m / s 50 \mathrm{ m/s} 50 m/s 40 ∘ 40^\circ 4 0 ∘
Two trains approach each other on parallel tracks. Train A travels at 60 m / s 60 \mathrm{ m/s} 60 m/s 40 m / s 40 \mathrm{ m/s} 40 m/s 80 m / s 80 \mathrm{ m/s} 80 m/s 2 k m 2 \mathrm{ km} 2 km
The position of a particle is given by x ( t ) = 2 t 3 − 9 t 2 + 12 t + 1 x(t) = 2t^3 - 9t^2 + 12t + 1 x ( t ) = 2 t 3 − 9 t 2 + 12 t + 1 t = 0 t = 0 t = 0 t = 4 s t = 4 \mathrm{ s} t = 4 s
Practice Problems Question 1: Two-object kinematics with relative motion A ball is thrown vertically upward from ground level with initial velocity 20 m / s 20 \mathrm{ m/s} 20 m/s 50 m 50 \mathrm{ m} 50 m g = 9.8 m / s 2 g = 9.8 \mathrm{ m/s^2} g = 9.8 m/ s 2
Answer Ball 1 (thrown up at t = 0 t = 0 t = 0 y 1 = 20 t − 4.9 t 2 y_1 = 20t - 4.9t^2 y 1 = 20 t − 4.9 t 2
Ball 2 (dropped at t = 1 t = 1 t = 1 y 2 = 50 − 4.9 ( t − 1 ) 2 y_2 = 50 - 4.9(t - 1)^2 y 2 = 50 − 4.9 ( t − 1 ) 2 t ≥ 1 t \ge 1 t ≥ 1
Set y 1 = y 2 y_1 = y_2 y 1 = y 2 20 t − 4.9 t 2 = 50 − 4.9 ( t 2 − 2 t + 1 ) = 50 − 4.9 t 2 + 9.8 t − 4.9 = 45.1 − 4.9 t 2 + 9.8 t 20t - 4.9t^2 = 50 - 4.9(t^2 - 2t + 1) = 50 - 4.9t^2 + 9.8t - 4.9 = 45.1 - 4.9t^2 + 9.8t 20 t − 4.9 t 2 = 50 − 4.9 ( t 2 − 2 t + 1 ) = 50 − 4.9 t 2 + 9.8 t − 4.9 = 45.1 − 4.9 t 2 + 9.8 t
20 t = 45.1 + 9.8 t 20t = 45.1 + 9.8t 20 t = 45.1 + 9.8 t 10.2 t = 45.1 10.2t = 45.1 10.2 t = 45.1 t = 4.42 s t = 4.42 \mathrm{ s} t = 4.42 s
Height: y 1 = 20 ( 4.42 ) − 4.9 ( 4.42 ) 2 = 88.4 − 95.7 = − 7.3 m y_1 = 20(4.42) - 4.9(4.42)^2 = 88.4 - 95.7 = -7.3 \mathrm{ m} y 1 = 20 ( 4.42 ) − 4.9 ( 4.42 ) 2 = 88.4 − 95.7 = − 7.3 m
This is negative, meaning the balls don’t meet before ball 1 hits the ground. Ball 1 lands when y 1 = 0 y_1 = 0 y 1 = 0 20 t − 4.9 t 2 = 0 20t - 4.9t^2 = 0 20 t − 4.9 t 2 = 0 t = 20 / 4.9 = 4.08 s t = 20/4.9 = 4.08 \mathrm{ s} t = 20/4.9 = 4.08 s
Since 4.42 > 4.08 4.42 > 4.08 4.42 > 4.08
Question 2: Projectile motion with angle optimization A football is kicked from ground level with speed 25 m / s 25 \mathrm{ m/s} 25 m/s
Answer For maximum range on level ground, the launch angle is 45 ∘ 45^\circ 4 5 ∘
Range: R = v 2 sin ( 2 θ ) g = 25 2 sin ( 90 ∘ ) 9.8 = 625 9.8 = 63.8 m R = \frac{v^2 \sin(2\theta)}{g} = \frac{25^2 \sin(90^\circ)}{9.8} = \frac{625}{9.8} = 63.8 \mathrm{ m} R = g v 2 s i n ( 2 θ ) = 9.8 2 5 2 s i n ( 9 0 ∘ ) = 9.8 625 = 63.8 m
Time of flight: T = 2 v sin θ g = 2 × 25 × sin ( 45 ∘ ) 9.8 = 35.36 9.8 = 3.61 s T = \frac{2v \sin\theta}{g} = \frac{2 \times 25 \times \sin(45^\circ)}{9.8} = \frac{35.36}{9.8} = 3.61 \mathrm{ s} T = g 2 v s i n θ = 9.8 2 × 25 × s i n ( 4 5 ∘ ) = 9.8 35.36 = 3.61 s
Question 3: Kinematics with calculus The velocity of a particle is given by v ( t ) = 6 t 2 − 18 t + 12 m / s v(t) = 6t^2 - 18t + 12 \mathrm{ m/s} v ( t ) = 6 t 2 − 18 t + 12 m/s 0 ≤ t ≤ 4 s 0 \le t \le 4 \mathrm{ s} 0 ≤ t ≤ 4 s
Answer Set v = 0 v = 0 v = 0 6t^2 - 18t + 12 = 0$$t^2 - 3t + 2 = 0$$(t-1)(t-2) = 0 . Direction changes at t = 1 t = 1 t = 1 t = 2 t = 2 t = 2
Acceleration: a = d v / d t = 12 t − 18 a = dv/dt = 12t - 18 a = d v / d t = 12 t − 18
Displacement = ∫ 0 4 v d t = [ 2 t 3 − 9 t 2 + 12 t ] 0 4 = 128 − 144 + 48 = 32 m \int_0^4 v \, dt = [2t^3 - 9t^2 + 12t]_0^4 = 128 - 144 + 48 = 32 \mathrm{ m} ∫ 0 4 v d t = [ 2 t 3 − 9 t 2 + 12 t ] 0 4 = 128 − 144 + 48 = 32 m
Total distance = ∫ 0 1 v d t + ∣ ∫ 1 2 v d t ∣ + ∫ 2 4 v d t \int_0^1 v \, dt + |\int_1^2 v \, dt| + \int_2^4 v \, dt ∫ 0 1 v d t + ∣ ∫ 1 2 v d t ∣ + ∫ 2 4 v d t
∫ 0 1 = [ 2 − 9 + 12 ] = 5 \int_0^1 = [2 - 9 + 12] = 5 ∫ 0 1 = [ 2 − 9 + 12 ] = 5 ∫ 1 2 = [ 16 − 36 + 24 ] − [ 2 − 9 + 12 ] = 4 − 5 = − 1 \int_1^2 = [16 - 36 + 24] - [2 - 9 + 12] = 4 - 5 = -1 ∫ 1 2 = [ 16 − 36 + 24 ] − [ 2 − 9 + 12 ] = 4 − 5 = − 1 ∫ 2 4 = [ 128 − 144 + 48 ] − [ 16 − 36 + 24 ] = 32 − 44 = − 12 \int_2^4 = [128 - 144 + 48] - [16 - 36 + 24] = 32 - 44 = -12 ∫ 2 4 = [ 128 − 144 + 48 ] − [ 16 − 36 + 24 ] = 32 − 44 = − 12
Total distance = 5 + 1 + 12 = 18 m 5 + 1 + 12 = 18 \mathrm{ m} 5 + 1 + 12 = 18 m
Displacement = 32 m, total distance = 18 m.
Question 4: Free fall with air resistance (conceptual) Explain qualitatively how the motion of a falling object differs with and without air resistance. Sketch velocity-time graphs for both cases and label the terminal velocity.
Answer Without air resistance: Velocity increases linearly with time (v = g t v = gt v = g t
With air resistance: As the object falls, air resistance (drag) increases with speed. Net force = weight - drag decreases. Acceleration decreases. Eventually, drag equals weight (net force = 0), and the object reaches terminal velocity (v T v_T v T v T v_T v T
Terminal velocity depends on mass, cross-sectional area, drag coefficient, and air density. A heavier object with smaller area reaches a higher terminal velocity.
Question 5: Relative motion in two dimensions A boat can travel at 5 m / s 5 \mathrm{ m/s} 5 m/s 100 m 100 \mathrm{ m} 100 m 3 m / s 3 \mathrm{ m/s} 3 m/s
Answer Heading directly across: Time to cross = 100 / 5 = 20 s 100 / 5 = 20 \mathrm{ s} 100/5 = 20 s 3 × 20 = 60 m 3 \times 20 = 60 \mathrm{ m} 3 × 20 = 60 m
To land directly across: The boat must angle upstream so its upstream component cancels the current. Let θ \theta θ
5 sin θ = 3 5\sin\theta = 3 5 sin θ = 3 \sin\theta = 3/5 = 0.6$$\theta = 36.9^\circ .
Velocity across river: 5 cos θ = 5 × 0.8 = 4 m / s 5\cos\theta = 5 \times 0.8 = 4 \mathrm{ m/s} 5 cos θ = 5 × 0.8 = 4 m/s 100 / 4 = 25 s 100 / 4 = 25 \mathrm{ s} 100/4 = 25 s
Cross-References Newton’s Laws — Forces cause acceleration, connecting kinematic descriptions of motion to the dynamics that explain why objects move.Work, Energy, and Power — Kinematic equations derive from energy conservation, linking motion analysis to energy methods.Momentum and Impulse — Impulse equals change in momentum, connecting force-time analysis to the kinematic outcomes.AP Calculus, Derivatives :References Young, H. D. & Freedman, R. A. (2019). University Physics with Modern Physics (15th ed.). Pearson. Halliday, D., Resnick, R. & Walker, J. (2013). Fundamentals of Physics (10th ed.). John Wiley & Sons. Serway, R. A. & Jewett, J. W. (2018). Physics for Scientists and Engineers with Modern Physics (10th ed.). Cengage Learning. Hewitt, P. G. (2015). Conceptual Physics (12th ed.). Pearson. Giancoli, D. C. (2013). Physics: Principles with Applications (7th ed.). Pearson. 