Kinematics

Board Coverage

Board	Paper	Notes
AQA	Paper 1	1D kinematics, projectiles
Edexcel	P1	Similar
OCR (A)	Paper 1	Includes variable acceleration
CIE (9709)	P1, P4	1D in P1; 2D/projectiles in P4

:::info The formula booklet lists the SUVAT equations. You must be able to select the correct Equation and apply it to multi-stage problems. :::

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1. Fundamental Quantities

1.1 Definitions

• Displacement $s$: position relative to a reference point (vector, measured in metres, m).
• Velocity $v$: rate of change of displacement (vector, m/s).
• Speed: magnitude of velocity (scalar, m/s).
• Acceleration $a$: rate of change of velocity (vector, m/s$^2$).

1.2 Relationships via calculus

$v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

$s = \int v\,dt, \qquad v = \int a\,dt$

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2. The SUVAT Equations

2.1 Derivation from calculus

Assuming constant acceleration $a$:

Start from the definition of acceleration:

$a = \frac{dv}{dt}$

Since $a$ is constant, integrate both sides with respect to $t$:

$\int a\,dt = \int \frac{dv}{dt}\,dt \implies at + C_1 = v$

Applying the initial condition $v = u$ when $t = 0$: $C_1 = u$.

$\boxed{v = u + at} \quad \mathrm{(Equation 2)}$

Now use $v = ds/dt$:

$\frac{ds}{dt} = u + at$

Integrate with respect to $t$:

$s = \int (u + at)\,dt = ut + \frac{1}{2}at^2 + C_2$

Since $s = 0$ when $t = 0$: $C_2 = 0$.

$\boxed{s = ut + \tfrac{1}{2}at^2} \quad \mathrm{(Equation 1)}$

Eliminating $t$ from (1) and (2): $t = (v-u)/a$.

$s = u\frac{v-u}{a} + \frac{1}{2}a\frac{(v-u)^2}{a^2} = \frac{uv - u^2}{a} + \frac{v^2 - 2uv + u^2}{2a}$

$s = \frac{2uv - 2u^2 + v^2 - 2uv + u^2}{2a} = \frac{v^2 - u^2}{2a}$

$\boxed{v^2 = u^2 + 2as} \quad \mathrm{(Equation 3)}$

From (1): $s = ut + \tfrac{1}{2}at^2 = \tfrac{1}{2}(2u + at)t = \tfrac{1}{2}(u + u + at)t$.

Using $v = u + at$: $\boxed{s = \tfrac{1}{2}(u+v)t} \quad \mathrm{(Equation 4)}$

Eliminating $a$ from (1) and (2): $a = (v-u)/t$.

$s = ut + \tfrac{1}{2}\frac{v-u}{t}t^2 = ut + \tfrac{1}{2}(v-u)t = \tfrac{1}{2}(u+v)t$. (Same as Equation 4.)

$s = \tfrac{1}{2}(u+v)t \implies v = \frac{2s}{t} - u \quad \mathrm{(useful when } a \mathrm{ is unknown)}$

:::tip The key insight is that all five SUVAT equations follow from two integrations of the Constant acceleration $a = dv/dt$Using the initial conditions $v(0) = u$ and $s(0) = 0$. Memorising the derivation means you can recover any formula you forget. :::

2.2 Summary

Equation	Variables	Missing
$v = u + at$	$v, u, a, t$	$s$
$s = ut + \frac{1}{2}at^2$	$s, u, a, t$	$v$
$v^2 = u^2 + 2as$	$v, u, a, s$	$t$
$s = \frac{1}{2}(u+v)t$	$s, u, v, t$	$a$

:::tip Always write down what you know and what you need. Choose the equation that contains all Known quantities and the unknown. Draw a diagram with positive direction marked. :::

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3. Motion Graphs

3.1 Displacement-time graphs

• Gradient = velocity
• Horizontal line = stationary
• Curved line = acceleration
• Area under the graph has no direct meaning

3.2 Velocity-time graphs

• Gradient = acceleration
• Area under the graph = displacement
• Horizontal line = constant velocity
• Straight line through origin = constant acceleration from rest

3.3 Acceleration-time graphs

• Area under the graph = change in velocity

3.4 Interpreting displacement-time graphs in detail

The gradient of an $s$-$t$ graph gives the velocity at that instant. The sign of the gradient Tells you the direction of motion, and the steepness tells you the speed.

• Positive, increasing gradient: particle moves in the positive direction with increasing speed (positive acceleration).
• Positive, decreasing gradient: particle moves in the positive direction but is decelerating.
• Zero gradient (horizontal): particle is instantaneously at rest ($v = 0$). This may be a turning point.
• Negative gradient: particle moves in the negative direction.
• Concave-up curve ($\frac{d^2s}{dt^2} \gt 0$): acceleration is positive.
• Concave-down curve ($\frac{d^2s}{dt^2} \lt 0$): acceleration is negative.

A common mistake is assuming a particle is at rest only when $s = 0$. In fact, the particle is at Rest whenever the gradient is zero, regardless of the displacement value.

3.5 Interpreting velocity-time graphs in detail

The area between the $v$-$t$ curve and the $t$-axis gives the displacement (with sign). To find The total distance travelled, you must take the absolute value of velocity in each region before Integrating, or equivalently add the magnitudes of the areas above and below the axis.

• Area above the $t$-axis: displacement in the positive direction.
• Area below the $t$-axis: displacement in the negative direction.
• Total distance = (area above) + |area below|.

The gradient of the tangent to a $v$-$t$ curve gives the instantaneous acceleration. For a Straight-line $v$-$t$ graph, the acceleration is constant and equals the gradient of that line.

:::caution Displacement $\neq$ distance. If a particle moves $10\,\mathrm{m}$ forwards and then $10\,\mathrm{m}$ backwards, its displacement is $0$ but its total distance is $20\,\mathrm{m}$. On a $v$-$t$ graph, the signed areas cancel (displacement), but the unsigned areas add (distance). :::

3.6 Worked example: graphs

A particle moves so that its displacement $s$ metres from a fixed point $O$ at time $t$ seconds is Given by $s = t^3 - 9t^2 + 24t$.

The velocity is $v = ds/dt = 3t^2 - 18t + 24 = 3(t^2 - 6t + 8) = 3(t-2)(t-4)$.

• $v = 0$ at $t = 2$ and $t = 4$: the particle is instantaneously at rest at these times.
• For $0 \lt t \lt 2$: $v \gt 0$ (moving in positive direction).
• For $2 \lt t \lt 4$: $v \lt 0$ (moving in negative direction — it has reversed).
• For $t \gt 4$: $v \gt 0$ (moving in positive direction again).

The acceleration is $a = dv/dt = 6t - 18 = 6(t - 3)$.

• At $t = 3$: $a = 0$The particle changes from decelerating to accelerating (in the positive sense).

Displacement at key times: s(0) = 0$$s(2) = 8 - 36 + 48 = 20$$s(4) = 64 - 144 + 96 = 16 $s(6) = 216 - 324 + 144 = 36$.

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4. Projectiles

4.1 Assumptions

• Motion is in a vertical plane.
• The only force is gravity (no air resistance).
• Horizontal and vertical motions are independent.

4.2 Horizontal motion

$x = v\cos\theta \cdot t, \quad a_x = 0$

Since there is no horizontal acceleration, the horizontal velocity $v_x = v\cos\theta$ remains Constant throughout the flight.

4.3 Vertical motion

$y = v\sin\theta \cdot t - \frac{1}{2}gt^2, \quad v_y = v\sin\theta - gt$

4.4 Derivation of the trajectory equation

From horizontal: $t = \dfrac◆LB◆x◆RB◆◆LB◆v\cos\theta◆RB◆$.

Substitute into vertical:

$y = v\sin\theta \cdot \frac◆LB◆x◆RB◆◆LB◆v\cos\theta◆RB◆ - \frac{1}{2}g\left(\frac◆LB◆x◆RB◆◆LB◆v\cos\theta◆RB◆\right)^2$

$\boxed{y = x\tan\theta - \frac◆LB◆gx^2◆RB◆◆LB◆2v^2\cos^2\theta◆RB◆}$

This is a parabola — all projectile trajectories are parabolic (under constant gravity, no air Resistance).

4.5 Maximum height

At maximum height, $v_y = 0$:

$0 = v\sin\theta - gt_{\max} \implies t_{\max} = \frac◆LB◆v\sin\theta◆RB◆◆LB◆g◆RB◆$

$H_{\max} = \frac◆LB◆(v\sin\theta)^2◆RB◆◆LB◆2g◆RB◆$

4.6 Range

Time of flight: $y = 0 \implies t = \dfrac◆LB◆2v\sin\theta◆RB◆◆LB◆g◆RB◆$.

$R = v\cos\theta \cdot \frac◆LB◆2v\sin\theta◆RB◆◆LB◆g◆RB◆ = \frac◆LB◆v^2\sin 2\theta◆RB◆◆LB◆g◆RB◆$

Maximum range occurs when $\sin 2\theta = 1$I.e., $\theta = 45^\circ$.

$R_{\max} = \frac{v^2}{g}$

4.7 Velocity at any point on the trajectory

At any time $t$The velocity vector is:

$\mathbf{v} = \begin{pmatrix} v\cos\theta \\ v\sin\theta - gt \end{pmatrix}$

The speed at time $t$ is:

$|\mathbf{v}| = \sqrt◆LB◆(v\cos\theta)^2 + (v\sin\theta - gt)^2◆RB◆ = \sqrt◆LB◆v^2 - 2vgt\sin\theta + g^2t^2◆RB◆$

The angle the velocity makes with the horizontal at time $t$ is:

$\alpha = \arctan\left(\frac◆LB◆v\sin\theta - gt◆RB◆◆LB◆v\cos\theta◆RB◆\right)$

At the highest point ($t = v\sin\theta / g$), the velocity is purely horizontal: $\mathbf{v} = (v\cos\theta,\, 0)$. The speed at the highest point equals the horizontal component $v\cos\theta$.

On landing ($t = 2v\sin\theta / g$), the vertical component is $-v\sin\theta$So the speed equals The initial speed $v$. The landing angle with the horizontal equals the launch angle $\theta$ (by Symmetry).

4.8 Time to reach a given height

Setting $y = h$ and solving for $t$:

$h = v\sin\theta \cdot t - \frac{1}{2}gt^2 \implies \frac{1}{2}gt^2 - v\sin\theta \cdot t + h = 0$

$t = \frac◆LB◆v\sin\theta \pm \sqrt{(v\sin\theta)^2 - 2gh}◆RB◆◆LB◆g◆RB◆$

• If $(v\sin\theta)^2 \gt 2gh$: two solutions — the projectile passes through height $h$ twice (on the way up and on the way down).
• If $(v\sin\theta)^2 = 2gh$: one solution — $h$ is the maximum height.
• If $(v\sin\theta)^2 \lt 2gh$: no real solution — the projectile never reaches height $h$.

4.9 Projectiles launched from a height

If a projectile is launched from height $H$ above ground level, set $y = -H$ at landing (taking Upwards as positive):

$-H = v\sin\theta \cdot t - \frac{1}{2}gt^2$

$\frac{1}{2}gt^2 - v\sin\theta \cdot t - H = 0$

$t = \frac◆LB◆v\sin\theta + \sqrt{(v\sin\theta)^2 + 2gH}◆RB◆◆LB◆g◆RB◆$

(We take the positive root since $t \gt 0$.)

The horizontal range is then $R = v\cos\theta \cdot t$.

:::tip When a projectile is launched from a height, the time of flight is longer than for level Ground, and the angle for maximum range is less than $45^\circ$ (derived via calculus, but Beyond the core syllabus). :::

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5. Variable Acceleration

When acceleration is not constant, the SUVAT equations do not apply. Instead, use calculus:

$v = \frac{ds}{dt}, \quad a = \frac{dv}{dt}$

$s = \int v\,dt, \quad v = \int a\,dt$

Use initial conditions to find constants of integration.

5.1 Finding velocity from acceleration

Given $a = f(t)$Integrate to find $v$:

$v = \int a\,dt = \int f(t)\,dt = F(t) + C$

Use the initial velocity $v(0) = u$ to find $C$.

5.2 Finding displacement from velocity

Given $v = g(t)$Integrate to find $s$:

$s = \int v\,dt = \int g(t)\,dt = G(t) + K$

Use the initial displacement $s(0) = s_0$ to find $K$.

5.3 Acceleration in terms of displacement or velocity

Sometimes acceleration is given as a function of $s$ or $v$Not $t$.

Case 1: $a = f(v)$.

Use the chain rule: $a = \dfrac{dv}{dt} = \dfrac{dv}{ds} \cdot \dfrac{ds}{dt} = v\dfrac{dv}{ds}$.

This gives a separable differential equation:

$f(v) = v\frac{dv}{ds} \implies \int ds = \int \frac{v}{f(v)}\,dv$

Case 2: $a = f(s)$.

Again using $a = v\,dv/ds$:

$v\,\frac{dv}{ds} = f(s) \implies \int v\,dv = \int f(s)\,ds$

$\frac{v^2}{2} = F(s) + C$

This is equivalent to the work-energy principle: $\tfrac{1}{2}mv^2 = \mathrm{work done}$.

5.4 Definite integration for distance and displacement

When finding displacement over a time interval $[t_1, t_2]$:

$\Delta s = \int_{t_1}^{t_2} v\,dt$

When finding total distance, you must account for changes in direction. Find when $v = 0$ (turning Points), split the integral at those times, and take absolute values:

$\mathrm{Distance} = \int_{t_1}^{t_2} |v|\,dt$

5.5 Worked example: variable acceleration

A particle moves in a straight line. At time $t$ seconds, its acceleration is $a = 6t - 4\,\mathrm{m/s}^2$. When $t = 0$The particle is at rest at the origin. Find:

(a) The velocity at time $t$:

$v = \int (6t - 4)\,dt = 3t^2 - 4t + C$

Since $v = 0$ when $t = 0$: $C = 0$So $v = 3t^2 - 4t$.

(b) When the particle is at rest:

$v = 0 \implies 3t^2 - 4t = 0 \implies t(3t - 4) = 0 \implies t = 0$ or $t = 4/3\,\mathrm{s}$.

(c) The displacement at time $t$:

$s = \int (3t^2 - 4t)\,dt = t^3 - 2t^2 + K$

Since $s = 0$ when $t = 0$: $K = 0$So $s = t^3 - 2t^2$.

(d) The total distance travelled in the first 3 seconds:

The particle reverses direction at $t = 4/3$.

$s(4/3) = (64/27) - 2(16/9) = 64/27 - 96/27 = -32/27\,\mathrm{m}$.

$s(3) = 27 - 18 = 9\,\mathrm{m}$.

Distance $= |s(4/3) - s(0)| + |s(3) - s(4/3)| = |-32/27| + |9 - (-32/27)| = 32/27 + 275/27 = 307/27 \approx 11.37\,\mathrm{m}$.

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Problem Set

Problem 1

A car accelerates from rest at $2\,\mathrm{m/s}^2$ for 8 seconds. Find the distance travelled.

Solution 1

$u = 0$, $a = 2$, $t = 8$. Using $s = ut + \tfrac{1}{2}at^2$:

$s = 0 + \tfrac{1}{2}(2)(64) = 64\,\mathrm{m}$.

If you get this wrong, revise: The SUVAT Equations — Section 2.

Problem 2

A ball is thrown vertically upwards at $15\,\mathrm{m/s}$. Find the maximum height and the time to return to the thrower's hand. Take $g = 9.8\,\mathrm{m/s}^2$.

Solution 2

At max height: $v = 0$. $v^2 = u^2 + 2as \implies 0 = 225 - 2(9.8)s \implies s = 225/19.6 \approx 11.48\,\mathrm{m}$.

Time up: $v = u - gt \implies 0 = 15 - 9.8t \implies t = 15/9.8 \approx 1.53\,\mathrm{s}$.

Total time (up and down): $2 \times 1.53 = 3.06\,\mathrm{s}$.

If you get this wrong, revise: Maximum Height — Section 4.5.

Problem 3

A projectile is launched at $30\,\mathrm{m/s}$ at an angle of $40^\circ$ above the horizontal. Find the range and maximum height. Take $g = 9.8\,\mathrm{m/s}^2$.

Solution 3

$v_x = 30\cos 40° \approx 22.98\,\mathrm{m/s}$, $v_y = 30\sin 40° \approx 19.28\,\mathrm{m/s}$.

$H_{\max} = \dfrac{(19.28)^2}{2(9.8)} = \dfrac{371.72}{19.6} \approx 18.97\,\mathrm{m}$.

$R = \dfrac◆LB◆30^2 \sin 80°◆RB◆◆LB◆9.8◆RB◆ = \dfrac◆LB◆900 \times 0.9848◆RB◆◆LB◆9.8◆RB◆ \approx 90.44\,\mathrm{m}$.

If you get this wrong, revise: Projectiles — Section 4.

Problem 4

A train decelerates uniformly from $25\,\mathrm{m/s}$ to $10\,\mathrm{m/s}$ over a distance of $200\,\mathrm{m}$. Find the deceleration and the time taken.

Solution 4

$u = 25$$v = 10$$s = 200$.

$v^2 = u^2 + 2as \implies 100 = 625 + 400a \implies a = -525/400 = -1.3125\,\mathrm{m/s}^2$.

$v = u + at \implies 10 = 25 - 1.3125t \implies t = 15/1.3125 \approx 11.43\,\mathrm{s}$.

If you get this wrong, revise: The SUVAT Equations — Section 2.

Problem 5

A particle moves with velocity $v = 3t^2 - 2t + 1$ m/s. Find the displacement after 3 seconds, given $s = 0$ at $t = 0$.

Solution 5

$s = \int_0^3 (3t^2 - 2t + 1)\,dt = \left[t^3 - t^2 + t\right]_0^3 = 27 - 9 + 3 = 21\,\mathrm{m}$.

If you get this wrong, revise: Variable Acceleration — Section 5.

Problem 6

Show that the maximum range of a projectile on level ground is achieved at $45^\circ$.

Solution 6

$R = \dfrac◆LB◆v^2 \sin 2\theta◆RB◆◆LB◆g◆RB◆$. To maximise: $\dfrac◆LB◆dR◆RB◆◆LB◆d\theta◆RB◆ = \dfrac◆LB◆2v^2 \cos 2\theta◆RB◆◆LB◆g◆RB◆ = 0 \implies \cos 2\theta = 0 \implies 2\theta = 90° \implies \theta = 45^\circ$.

$\dfrac◆LB◆d^2R◆RB◆◆LB◆d\theta^2◆RB◆ = -\dfrac◆LB◆4v^2 \sin 2\theta◆RB◆◆LB◆g◆RB◆ \lt 0$ at $\theta = 45^\circ$Confirming A maximum. $\blacksquare$

If you get this wrong, revise: Range — Section 4.6.

Problem 7

A stone is dropped from a cliff of height $80\,\mathrm{m}$. Find the time to hit the ground and the speed on impact. Take $g = 9.8\,\mathrm{m/s}^2$.

Solution 7

$s = \tfrac{1}{2}gt^2 \implies 80 = 4.9t^2 \implies t^2 = 80/4.9 \implies t \approx 4.04\,\mathrm{s}$.

$v = gt = 9.8(4.04) \approx 39.6\,\mathrm{m/s}$.

If you get this wrong, revise: The SUVAT Equations — Section 2.

Problem 8

A particle is projected from a point $O$ on horizontal ground. It passes through a point $P$ which is $10\,\mathrm{m}$ horizontally and $5\,\mathrm{m}$ vertically from $O$. If the initial speed is $15\,\mathrm{m/s}$Find the possible angles of projection.

Solution 8

Trajectory: $y = x\tan\theta - \dfrac◆LB◆gx^2◆RB◆◆LB◆2v^2\cos^2\theta◆RB◆$.

$5 = 10\tan\theta - \dfrac◆LB◆9.8(100)◆RB◆◆LB◆2(225)\cos^2\theta◆RB◆ = 10\tan\theta - \dfrac◆LB◆980◆RB◆◆LB◆450\cos^2\theta◆RB◆$.

Using $\sec^2\theta = 1 + \tan^2\theta$:

$5 = 10\tan\theta - \dfrac{980}{450}(1+\tan^2\theta)$.

Let $u = \tan\theta$: $5 = 10u - \dfrac{98}{45}(1+u^2) = 10u - \dfrac{98}{45} - \dfrac{98}{45}u^2$.

$225 = 450u - 98 - 98u^2 \implies 98u^2 - 450u + 323 = 0$.

$u = \dfrac◆LB◆450 \pm \sqrt{202500 - 126604}◆RB◆◆LB◆196◆RB◆ = \dfrac◆LB◆450 \pm \sqrt{75896}◆RB◆◆LB◆98◆RB◆ = \dfrac◆LB◆450 \pm 275.5◆RB◆◆LB◆196◆RB◆$.

$u \approx 3.702$ or $u \approx 0.890$.

$\theta \approx 74.9^\circ$ or $\theta \approx 41.7^\circ$.

If you get this wrong, revise: Trajectory Equation — Section 4.4.

Problem 9

A car travels at $20\,\mathrm{m/s}$ for 30 seconds, then decelerates at $1.5\,\mathrm{m/s}^2$ until it stops. Find the total distance and total time.

Solution 9

Phase 1: $s_1 = 20 \times 30 = 600\,\mathrm{m}$$t_1 = 30\,\mathrm{s}$.

Phase 2: $v = u + at \implies 0 = 20 - 1.5t \implies t = 40/3 \approx 13.33\,\mathrm{s}$.

$s_2 = ut + \tfrac{1}{2}at^2 = 20(40/3) - \tfrac{1}{2}(1.5)(1600/9) = 800/3 - 400/3 = 400/3 \approx 133.3\,\mathrm{m}$.

Total: s = 600 + 133.3 = 733.3\,\mathrm{m}$$t = 30 + 13.33 = 43.33\,\mathrm{s}.

If you get this wrong, revise: The SUVAT Equations — Section 2.

Problem 10

The velocity of a particle is given by $v = 6t - t^2$ for $0 \leq t \leq 6$. Find the maximum velocity and the total distance travelled.

Solution 10

$a = dv/dt = 6 - 2t = 0 \implies t = 3$. $v_{\max} = 18 - 9 = 9\,\mathrm{m/s}$.

Distance: $s = \int_0^6 (6t-t^2)\,dt = [3t^2 - t^3/3]_0^6 = 108 - 72 = 36\,\mathrm{m}$.

If you get this wrong, revise: Variable Acceleration — Section 5.

Problem 11

Two balls are dropped from the same height, the second $1\,\mathrm{s}$ after the first. How far apart are they when the first hits the ground (height $= 45\,\mathrm{m}$)?

Solution 11

First ball: $t = \sqrt{90/9.8} \approx 3.03\,\mathrm{s}$.

Second ball at $t = 3.03$: has been falling for $2.03\,\mathrm{s}$.

$s_2 = \tfrac{1}{2}(9.8)(2.03)^2 = 4.9 \times 4.121 \approx 20.19\,\mathrm{m}$.

Separation: $45 - 20.19 = 24.81\,\mathrm{m}$.

If you get this wrong, revise: The SUVAT Equations — Section 2.

Problem 12

A projectile is launched from ground level and just clears a wall $20\,\mathrm{m}$ high and $40\,\mathrm{m}$ away. If the launch angle is $50^\circ$Find the minimum launch speed.

Solution 12

$y = x\tan\theta - \dfrac◆LB◆gx^2◆RB◆◆LB◆2v^2\cos^2\theta◆RB◆$.

$20 = 40\tan 50° - \dfrac◆LB◆9.8 \times 1600◆RB◆◆LB◆2v^2\cos^2 50°◆RB◆$.

$20 = 40(1.1918) - \dfrac{15680}{2v^2(0.4132)} = 47.67 - \dfrac{15680}{0.8263v^2} = 47.67 - \dfrac{18976.9}{v^2}$.

$\dfrac{18976.9}{v^2} = 27.67 \implies v^2 = 686.0 \implies v \approx 26.2\,\mathrm{m/s}$.

If you get this wrong, revise: Trajectory Equation — Section 4.4.

Problem 13

A particle moves with acceleration $a = 4 - 2t\,\mathrm{m/s}^2$. When $t = 0$$v = 3\,\mathrm{m/s}$ and $s = 0$. Find the velocity and displacement when $t = 5$. Also find when the particle is at rest.

Solution 13

$v = \int (4 - 2t)\,dt = 4t - t^2 + C$. Since $v(0) = 3$: $C = 3$So $v = 4t - t^2 + 3$.

$s = \int (4t - t^2 + 3)\,dt = 2t^2 - t^3/3 + 3t + K$. Since $s(0) = 0$: $K = 0$So $s = 2t^2 - t^3/3 + 3t$.

At $t = 5$: $v = 20 - 25 + 3 = -2\,\mathrm{m/s}$ $s = 50 - 125/3 + 15 = 65 - 41.67 = 23.33\,\mathrm{m}$.

At rest: $v = 0 \implies -t^2 + 4t + 3 = 0 \implies t^2 - 4t - 3 = 0 \implies t = (4 \pm \sqrt{16+12})/2 = 2 \pm \sqrt{7}$.

$t = 2 + \sqrt{7} \approx 4.65\,\mathrm{s}$ (taking the positive root).

If you get this wrong, revise: Variable Acceleration — Section 5.

Problem 14

A projectile is launched from the top of a cliff $60\,\mathrm{m}$ high at $20\,\mathrm{m/s}$ horizontally. Find the time to hit the ground, the horizontal distance from the base of the cliff, and the speed on impact. Take $g = 9.8\,\mathrm{m/s}^2$.

Solution 14

Horizontal: $v_x = 20\,\mathrm{m/s}$ (constant). Vertical: $u_y = 0$$a_y = 9.8$$s_y = 60$ (downwards positive).

$s_y = \tfrac{1}{2}gt^2 \implies 60 = 4.9t^2 \implies t = \sqrt{60/4.9} \approx 3.50\,\mathrm{s}$.

Horizontal distance: $x = 20 \times 3.50 = 70.0\,\mathrm{m}$.

Vertical velocity on impact: $v_y = gt = 9.8 \times 3.50 = 34.3\,\mathrm{m/s}$.

Speed: $|\mathbf{v}| = \sqrt{20^2 + 34.3^2} = \sqrt{400 + 1176.49} = \sqrt{1576.49} \approx 39.7\,\mathrm{m/s}$.

If you get this wrong, revise: Projectiles from a Height — Section 4.9.

Problem 15

The velocity of a particle is $v = 2t^3 - 9t^2 + 12t - 5$ m/s. Find the total distance travelled between $t = 0$ and $t = 3$.

Solution 15

First find when $v = 0$: $2t^3 - 9t^2 + 12t - 5 = 0$.

Testing $t = 1$: $2 - 9 + 12 - 5 = 0$. So $(t-1)$ is a factor.

$2t^3 - 9t^2 + 12t - 5 = (t-1)(2t^2 - 7t + 5) = (t-1)(2t-5)(t-1) = (t-1)^2(2t-5)$.

So $v = 0$ at $t = 1$ and $t = 2.5$.

Check the sign of $v$: for $0 \lt t \lt 1$Test $t = 0.5$: $v = 0.25 - 2.25 + 6 - 5 = -1 \lt 0$. For $1 \lt t \lt 2.5$Test $t = 2$: $v = 16 - 36 + 24 - 5 = -1 \lt 0$. For $t \gt 2.5$Test $t = 3$: $v = 54 - 81 + 36 - 5 = 4 \gt 0$.

So $v \lt 0$ for $0 \lt t \lt 2.5$ and $v \gt 0$ for $t \gt 2.5$.

$s(2.5) = \int_0^{2.5} v\,dt = \left[\tfrac{1}{2}t^4 - 3t^3 + 6t^2 - 5t\right]_0^{2.5} = 19.531 - 46.875 + 37.5 - 12.5 = -2.344\,\mathrm{m}$.

$s(3) = \left[\tfrac{1}{2}t^4 - 3t^3 + 6t^2 - 5t\right]_0^3 = 40.5 - 81 + 54 - 15 = -1.5\,\mathrm{m}$.

Distance $= |s(2.5) - s(0)| + |s(3) - s(2.5)| = |-2.344| + |-1.5 - (-2.344)| = 2.344 + 0.844 = 3.188\,\mathrm{m}$.

If you get this wrong, revise: Definite Integration for Distance — Section 5.4.

Problem 16

A ball is thrown at $12\,\mathrm{m/s}$ at an angle of $60^\circ$ above the horizontal from a point $2\,\mathrm{m}$ above level ground. Find the speed and direction of the ball when it hits the ground. Take $g = 9.8\,\mathrm{m/s}^2$.

Solution 16

$v_x = 12\cos 60° = 6\,\mathrm{m/s}$$v_{y0} = 12\sin 60° = 6\sqrt{3} \approx 10.39\,\mathrm{m/s}$.

Taking upwards as positive with launch at $s_y = 2$:

$s_y = v_{y0}\,t - \tfrac{1}{2}gt^2 = 2$. On hitting ground: $s_y = 0$ (relative to ground).

$0 = 2 + 10.39t - 4.9t^2 \implies 4.9t^2 - 10.39t - 2 = 0$.

$t = \dfrac◆LB◆10.39 + \sqrt{107.95 + 39.2}◆RB◆◆LB◆9.8◆RB◆ = \dfrac◆LB◆10.39 + \sqrt{147.15}◆RB◆◆LB◆9.8◆RB◆ = \dfrac{10.39 + 12.13}{9.8} \approx 2.29\,\mathrm{s}$.

Vertical velocity at impact: $v_y = 10.39 - 9.8(2.29) = 10.39 - 22.44 = -12.05\,\mathrm{m/s}$.

Speed: $\sqrt{6^2 + 12.05^2} = \sqrt{36 + 145.20} = \sqrt{181.20} \approx 13.46\,\mathrm{m/s}$.

Angle below horizontal: $\arctan(12.05/6) \approx 63.5^\circ$.

If you get this wrong, revise: Velocity at Any Point — Section 4.7.

Problem 17

A particle moves so that $a = -6s\,\mathrm{m/s}^2$Where $s$ is the displacement from a fixed point. When $s = 0$$v = 8\,\mathrm{m/s}$. Find the velocity when $s = 1$.

Solution 17

Using $a = v\,dv/ds$:

$v\,\frac{dv}{ds} = -6s$

$\int v\,dv = \int -6s\,ds$

$\frac{v^2}{2} = -3s^2 + C$

When s = 0$$v = 8: $64/2 = C \implies C = 32$.

$\frac{v^2}{2} = -3s^2 + 32$

When $s = 1$: $v^2/2 = -3 + 32 = 29 \implies v^2 = 58 \implies v = \sqrt{58} \approx 7.62\,\mathrm{m/s}$.

The particle is still moving in the positive direction ($v \gt 0$) since it has not yet reached The turning point where $v = 0$ (which occurs at $s^2 = 32/3$I.e., $s \approx 3.27\,\mathrm{m}$).

If you get this wrong, revise: Acceleration in Terms of Displacement — Section 5.3.

Problem 18

A particle $P$ is projected from a point $A$ on horizontal ground with speed $u$ at an angle $\theta$ above the horizontal. At the instant $P$ passes through the highest point of its trajectory, a second particle $Q$ is projected vertically upwards from the point on the ground directly below that highest point. Given that $P$ and $Q$ collide, find an expression for the speed of projection of $Q$ in terms of $u$ and $\theta$.

Solution 18

Highest point of $P$'s trajectory: $x = \dfrac◆LB◆u^2\sin 2\theta◆RB◆◆LB◆2g◆RB◆$, $y = \dfrac◆LB◆u^2\sin^2\theta◆RB◆◆LB◆2g◆RB◆$At time $t_1 = \dfrac◆LB◆u\sin\theta◆RB◆◆LB◆g◆RB◆$.

After $t_1$, $P$ is in free fall with $v_y = 0$ at $t_1$So for $t \geq t_1$:

$y_P = \dfrac◆LB◆u^2\sin^2\theta◆RB◆◆LB◆2g◆RB◆ - \dfrac{1}{2}g(t - t_1)^2$.

$x_P = u\cos\theta \cdot t$.

For collision, $Q$ must be at the same $(x, y)$. Since $Q$ is projected vertically from directly Below the highest point, $Q$‘s horizontal position is always $x = u^2\sin 2\theta / (2g)$.

For $P$ to be at this $x$-coordinate at time $t$: $u\cos\theta \cdot t = u^2\sin 2\theta / (2g) = u^2\sin\theta\cos\theta / g$So $t = u\sin\theta / g = t_1$.

This means collision occurs at $t = t_1$The instant of the highest point. But $Q$ is projected at That instant, so for collision we need $y_Q(0^+) = y_P(t_1) = H$.

$Q$ starts at ground level ($y_Q = 0$) and must reach $y = H = u^2\sin^2\theta / (2g)$.

For $Q$: $v_Q = w - gt$, $y_Q = wt - \tfrac{1}{2}gt^2$Where $w$ is the projection speed.

Collision at $y = H$ when $t = 0$ is impossible ($Q$ starts at $y = 0$). So collision must occur at Some $\Delta t \gt 0$ after $t_1$.

At time $t_1 + \Delta t$:

$y_P = H - \tfrac{1}{2}g(\Delta t)^2$, $y_Q = w\,\Delta t - \tfrac{1}{2}g(\Delta t)^2$.

For collision: $H = w\,\Delta t$. Also, $x$ must match: $u\cos\theta(t_1 + \Delta t) = u^2\sin\theta\cos\theta/g + u\cos\theta\,\Delta t$. This is satisfied For all $\Delta t$ since $u\cos\theta \cdot t_1 = u^2\sin\theta\cos\theta/g$.

So any $w$ and $\Delta t$ with $w\,\Delta t = H$ gives a collision. The minimum speed is $w = H/\Delta t$ for $\Delta t \to 0^+$But in practice we need a finite time.

If we require collision at the highest point itself ($\Delta t \to 0$), then $w \to \infty$Which Is unphysical. The problem states they collide at some time after projection. Since no further Constraint is given, we take $w$ as a free parameter satisfying $w\,\Delta t = u^2\sin^2\theta / (2g)$ for some $\Delta t \gt 0$.

If you get this wrong, revise: Projectiles — Section 4.

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:::tip Tip Ready to test your understanding of Kinematics? The contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Kinematics with other topics to test synthesis under exam conditions.

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Common Pitfalls

1. Incorrectly applying $\vec{F} = m\vec{a}$ when forces are not collinear — resolve into components first.

2. Forgetting to include units in final answers, especially when working with derived units like $\text{N}\,\text{kg}^{-1}\,\text{m}^2$.

3. Rounding intermediate answers too early, which compounds errors in multi-step calculations.

4. Confusing displacement with distance, or velocity with speed, particularly in graphs and calculations.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.