M 2 Kinematics of motion in a straight line | A Level Maths 9709

Home › Learning Hub › A Level Maths › M 2: Kinematics of motion in a straight line

🎯What you need to be able to do

Distinguish distance from displacement, and speed from velocity.
Use the four constant-acceleration formulae, and choose the right one.
Solve problems involving vertical motion under gravity.
Interpret displacement–time and velocity–time graphs, using gradients and areas.
Use calculus when the acceleration is not constant.
Find total distance travelled when the direction of motion changes.

📚The mathematics

The vocabulary

Displacement is position relative to a starting point, with sign; distance is how far the body has actually travelled. Velocity is the rate of change of displacement and carries a sign; speed is its magnitude and never does.

The distinction matters most when a body reverses. A ball thrown up and caught again has zero displacement but has travelled a real distance, and its velocity changes sign at the top while its speed passes through zero.

Constant acceleration

\( v = u + at \)

\( s = ut + \tfrac{1}{2}at^{2} \)

\( v^{2} = u^{2} + 2as \)

\( s = \tfrac{1}{2}(u+v)t \)

These hold only when the acceleration is constant. Check that before writing any of them down — if the question gives acceleration as a function of time, you need calculus instead.

Each formula omits a different quantity: the first has no \(s\), the second no \(v\), the third no \(t\), the fourth no \(a\). List the three you know and the one you want, then pick the formula that leaves out the one you neither know nor need. That turns a four-way choice into a one-second decision.

Signs and direction

Fix a positive direction before you start, and stick to it. Everything pointing the other way is negative — including \(g\) if you take upwards as positive. Most kinematics errors are sign errors, and they are invisible in the arithmetic. Writing “taking upwards as positive” at the top of your working costs one line and prevents most of them.

Note that a negative acceleration does not mean “slowing down”. It means the acceleration points in the negative direction. A body moving in the negative direction with negative acceleration is speeding up. Drop the word deceleration and let the signs carry the meaning.

Vertical motion under gravity

With air resistance neglected, a body moving vertically has constant acceleration \(g\) downwards — at every instant, including at the highest point of its flight, where the velocity is momentarily zero but the acceleration certainly is not.

Two standard facts follow. At the highest point \( v = 0 \), which is usually the condition that unlocks the question. And for a body returning to its launch height, the displacement over the whole flight is zero, not twice the maximum height.

Motion graphs

On a displacement–time graph, the gradient is velocity.
On a velocity–time graph, the gradient is acceleration and the area between the line and the time axis is displacement.

Remember it as gradient going down the list, area coming back up. On a velocity–time graph the area is signed: area below the axis counts as negative displacement. For total distance, add the magnitudes of the areas instead.

Velocity–time graphs made of straight lines are the fastest route to many problems, since the area is just triangles and trapezia — often quicker than the formulae.

Variable acceleration: calculus

When the acceleration is not constant, differentiate going down and integrate coming back up:

\( v = \dfrac{\mathrm{d}s}{\mathrm{d}t} \)

\( a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}} \)

\( s = \displaystyle\int v\,\mathrm{d}t \)

\( v = \displaystyle\int a\,\mathrm{d}t \)

Integration introduces a constant, so you need an initial condition — usually the value at \( t = 0 \) — to pin it down. “Starts from rest” means \( v = 0 \) at \( t = 0 \); “starts at the origin” means \( s = 0 \) at \( t = 0 \). These are two different conditions and questions supply whichever they need.

For total distance when the velocity changes sign, find the times when \( v = 0 \), split the integral at each, and add the magnitudes — exactly as with areas in P1 8.

✏️Worked example

Take \( g = 10 \) m s\(^{-2}\). (a) A stone is thrown vertically upwards at 21 m s\(^{-1}\) from a point 2 m above the ground. Find the greatest height above the ground and the time until it hits the ground. (b) A particle moves in a straight line with velocity \( v = 3t^{2} - 12t + 9 \) m s\(^{-1}\) for \( 0 \le t \le 4 \). Find the total distance travelled.

(a) Take upwards as positive, so \( u = 21 \) and \( a = -10 \), measuring displacement from the throwing point.

At the greatest height \( v = 0 \). Using \( v^{2} = u^{2} + 2as \):

\[ 0 = 441 - 20s \;\Longrightarrow\; s = 22.05\ \text{m} \]

That is above the throwing point, which was already 2 m up, so the greatest height above the ground is \( 24.05 \) m.

The stone hits the ground when its displacement from the throwing point is \( -2 \) m. Using \( s = ut + \tfrac12 at^{2} \):

\[ -2 = 21t - 5t^{2} \;\Longrightarrow\; 5t^{2} - 21t - 2 = 0 \]

By the quadratic formula, \( t = \dfrac{21 \pm \sqrt{441 + 40}}{10} = \dfrac{21 \pm 21.932}{10} \), giving \( t = 4.29 \) or \( t = -0.093 \). Time cannot be negative, so \( t = 4.29 \) s.

(b) First find where the velocity is zero: \( 3t^{2} - 12t + 9 = 3(t^{2} - 4t + 3) = 3(t-1)(t-3) = 0 \), so \( t = 1 \) and \( t = 3 \), both inside the interval. The particle therefore reverses twice, and the integral must be split.

Integrating, \( s = t^{3} - 6t^{2} + 9t \). Evaluating at the key times:

\( s(0) = 0 \)

\( s(1) = 1 - 6 + 9 = 4 \)

\( s(3) = 27 - 54 + 27 = 0 \)

\( s(4) = 64 - 96 + 36 = 4 \)

So the particle moves \( +4 \) m, then back \( -4 \) m, then forward \( +4 \) m again. The total distance is

\[ |4 - 0| + |0 - 4| + |4 - 0| = 12\ \text{m} \]

Check it. The net displacement over the whole interval is \( s(4) - s(0) = 4 \) m, which is much less than the 12 m travelled — exactly what you expect when the particle doubles back. If your distance and displacement come out equal, you have almost certainly missed a change of direction. In (a), the negative root \( t = -0.093 \) is not a mistake: it is where the stone would have been at ground level had it been launched from below, and discarding it with that reason is the right move.

Measure displacement consistently from one origin. In (a) the greatest height from the throwing point is 22.05 m, but the question asked for the height above the ground, so the 2 m must be added; meanwhile the landing condition is \( s = -2 \), not \( s = 0 \). Mixing the two origins — using \( s = 0 \) for landing but adding 2 for the height — is the standard error and produces a plausible-looking wrong time.

📝Practise

Work through these, then reveal the answer. Take \( g = 10 \) m s\(^{-2}\) throughout.

1. A car accelerates uniformly from 8 m s\(^{-1}\) to 26 m s\(^{-1}\) over 120 m. Find the acceleration and the time taken.

Known: \( u = 8 \), \( v = 26 \), \( s = 120 \); wanted: \(a\). The formula without \(t\) is \( v^{2} = u^{2} + 2as \): \( 676 = 64 + 240a \), so \( 240a = 612 \) and \( a = 2.55 \) m s\(^{-2}\). For the time, the formula without \(a\) is quickest: \( 120 = \tfrac12(8+26)t = 17t \), so \( t = 7.06 \) s.

2. A ball is dropped from rest at a height of 45 m. Find the time to reach the ground and the speed on impact.

Taking downwards as positive: \( u = 0 \), \( a = 10 \), \( s = 45 \). From \( s = ut + \tfrac12 at^{2} \): \( 45 = 5t^{2} \), so \( t^{2} = 9 \) and \( t = 3 \) s. Then \( v = u + at = 30 \) m s\(^{-1}\). (Or in one step, \( v = \sqrt{2as} = \sqrt{900} = 30 \).)

3. A velocity–time graph rises linearly from 0 to 15 m s\(^{-1}\) over 4 s, stays constant for 6 s, then falls linearly to 0 over 5 s. Find the total distance and the average speed.

The area is a trapezium. Triangle 1: \( \tfrac12(4)(15) = 30 \) m. Rectangle: \( 15 \times 6 = 90 \) m. Triangle 2: \( \tfrac12(5)(15) = 37.5 \) m. Total \( = 157.5 \) m over \( 4 + 6 + 5 = 15 \) s, so the average speed is \( \dfrac{157.5}{15} = 10.5 \) m s\(^{-1}\).

4. A particle has acceleration \( a = 6t - 4 \) m s\(^{-2}\) and velocity 5 m s\(^{-1}\) when \( t = 0 \). Find \(v\) when \( t = 3 \).

Integrate: \( v = 3t^{2} - 4t + c \). At \( t = 0 \), \( v = 5 \), so \( c = 5 \) and \( v = 3t^{2} - 4t + 5 \). At \( t = 3 \): \( 27 - 12 + 5 = 20 \) m s\(^{-1}\). Note the constant-acceleration formulae are unusable here — the acceleration depends on \(t\).

5. A stone is projected vertically upwards at 24 m s\(^{-1}\). For how long is it more than 20 m above its point of projection?

Taking upwards as positive, \( s = 24t - 5t^{2} \). Setting \( s = 20 \): \( 5t^{2} - 24t + 20 = 0 \), so \( t = \dfrac{24 \pm \sqrt{576 - 400}}{10} = \dfrac{24 \pm 13.266}{10} \), giving \( t = 1.073 \) and \( t = 3.727 \). The stone is above 20 m between these times, so for \( 3.727 - 1.073 = 2.65 \) s. Two roots are expected here — one going up, one coming down.

6. A particle moves with \( v = 4 - t^{2} \) m s\(^{-1}\) for \( 0 \le t \le 3 \). Find the displacement and the total distance travelled.

Displacement: \( \displaystyle\int_0^3 \left(4 - t^{2}\right)\mathrm{d}t = \left[4t - \tfrac{t^{3}}{3}\right]_0^3 = 12 - 9 = 3 \) m. For distance, note \( v = 0 \) at \( t = 2 \). From 0 to 2: \( 8 - \tfrac83 = \tfrac{16}{3} \) m. From 2 to 3: \( \left(12 - 9\right) - \tfrac{16}{3} = 3 - \tfrac{16}{3} = -\tfrac{7}{3} \) m. Total distance \( = \tfrac{16}{3} + \tfrac{7}{3} = \tfrac{23}{3} = 7.67 \) m, against a displacement of only 3 m.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

The Physics Classroom — kinematics and motion graphs
PhET — The Moving Man, for linking motion to its graphs in real time
Desmos — plot \( v(t) \) and shade the area to see displacement versus distance