Mathematics 9709/41 — May/June 2022

Cambridge A-Level · Mechanics · worked solutions for every part, with the mark scheme

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Topics Kinematics of Motion in a Straight Line · Newton's Laws of Motion · Forces and Equilibrium · Energy, Work and Power · Momentum

Q1Kinematics of Motion in a Straight LineFree sample

A car starts from rest and moves in a straight line with constant acceleration for a distance of $200\text{ m}$ , reaching a speed of $25\text{ m s}^{-1}$ . The car then travels at this speed for $400\text{ m}$ , before decelerating uniformly to rest over a period of $5\text{ s}$ .

(a)

Find the time for which the car is accelerating.

DifficultyEasy

Worked solution

Approach

Use the constant acceleration formula that involves $s$ , $u$ , $v$ and $t$ but not $a$ , since the acceleration is not given.

Working

For the accelerating stage, $u = 0$ , $v = 25\text{ m s}^{-1}$ and $s = 200\text{ m}$ . Therefore

s = \frac{u+v}{2}\,t

200 = \frac{0+25}{2}\,t = 12.5t.

Solving for $t$ :

t = \frac{200}{12.5} = 16\text{ s}.

(Equivalently, $v^2 = u^2 + 2as$ gives $a = 1.5625\text{ m s}^{-2}$ , and then $v = u + at$ also gives $t = 16\text{ s}$ .)

Answer

t = 16\text{ s}

Final answer

16 s

Detailed explanation

Walkthrough

The car accelerates from rest, so $u=0$ , to a final speed $v=25\text{ m s}^{-1}$ , over a distance $s=200\text{ m}$ . The acceleration is constant but its value is not given, so the most direct formula is $s = \frac{(u+v)}{2}t$ , which uses only $s,u,v,t$ . Substitute the known values to get $200 = 12.5t$ , then divide both sides by $12.5$ to find $t=16$ seconds.

Key Takeaways

This part tests choosing an appropriate suvat equation when one quantity, the acceleration, is unknown. The formula $s = \frac{(u+v)}{2}t$ is useful because it uses the average velocity and the time.

Common Mistakes

Using $v = u + at$ directly without first finding $a$ can lead to two unknowns.
Confusing the initial and final velocities: here $u=0$ and $v=25$ .
Swapping the distance and speed values when substituting.

Things to Be Careful About

The units must be consistent: distance in metres and speed in m/s, so time comes out in seconds. The answer is exactly $16$ s; no rounding is needed.

Techniques used

select the suvat formula linking displacement, initial velocity, final velocity and timesubstitute known valuessolve for time

(b)

Sketch the velocity–time graph for the motion of the car, showing the key points.

DifficultyMedium-Easy

Worked solution

Approach

Plot $v$ against $t$ . First find the time for the constant-speed stage, then use the known acceleration and deceleration intervals to locate all key points.

Working

The accelerating stage lasts $t=16$ s from part (a), so the point $(16,25)$ marks the end of acceleration.

During the constant-speed stage the car travels $400$ m at $25\text{ m s}^{-1}$ , so the time is

\frac{400}{25} = 16\text{ s}.

Therefore the constant-speed stage ends at

t = 16 + 16 = 32\text{ s}.

The car then decelerates to rest over $5$ s, ending at

t = 32 + 5 = 37\text{ s}.

The key points are $(0,0)$ , $(16,25)$ , $(32,25)$ and $(37,0)$ . The graph is a trapezium with a horizontal top edge.

Answer

Sketch the required trapezium on axes labelled $v$ and $t$ , with the key points above.

Final answer

Trapezium with key points (0,0), (16,25), (32,25), (37,0)

Detailed explanation

Walkthrough

The question requires a velocity-time graph. Start from rest at the origin $(0,0)$ . From part (a), acceleration lasts 16 s and reaches 25 m/s; this gives a rising straight segment to $(16,25)$ . The next stage is constant speed, so the graph is horizontal. Since the car moves 400 m at 25 m/s, this stage lasts $400/25=16$ s, from $t=16$ to $t=32$ . Finally, it decelerates uniformly to rest over 5 s, so the line falls from $(32,25)$ to $(37,0)$ . Join these with straight line segments. The result is a trapezium. The area under the graph equals the total distance, although here only the shape and coordinates are required.

Key Takeaways

On a velocity-time graph, constant velocity appears as a horizontal line; constant acceleration appears as a straight line with constant slope; deceleration to rest is a straight line to the $t$ -axis. The key coordinates combine the durations of each stage.

Common Mistakes

Forgetting to compute the duration of the constant-speed stage; it is not automatically 16 s.
Drawing the graph as a triangle only, missing the horizontal section.
Ending the graph at $(32,0)$ instead of continuing the deceleration to $t=37$ .
Not labelling the axes $v$ and $t$ .

Things to Be Careful About

Use the follow-through value of $t$ from part (a) if it differs. The final coordinate is $t=37$ , not $32$ ; the braking stage of 5 s must be added to the time at the start of braking. The horizontal segment must be at $v=25$ .

Techniques used

break the motion into three stagescompute the duration of the constant-speed stageplot the key coordinates on a velocity-time graph

(c)

Find the average speed of the car during its motion.

DifficultyMedium-Easy

Worked solution

Approach

Find the total time by adding the times of the three stages. Find the total distance by adding the distances of the three stages. Then divide total distance by total time to get average speed.

Working

From part (a), the acceleration takes $16$ s. The constant-speed stage takes

\frac{400}{25} = 16\text{ s},

and the braking stage takes $5$ s. Hence total time is

16 + 16 + 5 = 37\text{ s}.

Distance in the braking stage is the area of the final triangle on the velocity-time graph:

\frac{1}{2} \times 25 \times 5 = 62.5\text{ m}.

Total distance

200 + 400 + 62.5 = 662.5\text{ m}.

Average speed is total distance divided by total time, so

\text{average speed} = \frac{662.5}{37} \approx 17.9\text{ m s}^{-1} \ (3\text{ s.f.}).

(Equivalently, the area under the whole trapezium is $\frac{1}{2} \times 25 \times (16+37)=662.5$ m.)

Answer

\text{average speed} = 17.9\text{ m s}^{-1} \ (3\text{ s.f.})

Final answer

17.9 m/s (3 s.f.)

Detailed explanation

Walkthrough

Average speed is not the average of the three speeds; it is total distance divided by total time. We already know the acceleration stage lasts 16 s from part (a). The constant speed stage lasts $400/25=16$ s. The final stage lasts 5 s, so total time is $37$ s. For distance, the first stage is 200 m, the second is 400 m, and the final stage is the triangular area under the velocity-time graph: $\frac{1}{2}(25)(5)=62.5$ m. Total distance is therefore $662.5$ m. Finally divide $662.5$ by $37$ to obtain $17.9$ m/s.

Key Takeaways

The area under a velocity-time graph represents distance or displacement. A triangular segment can be used even without suvat. Average speed is a scalar quantity equal to total distance over total time.

Common Mistakes

Forgetting that the car still travels during the braking stage; the deceleration distance $62.5$ m must be included.
Using total time $32$ s instead of $37$ s.
Averaging the speeds $0,25,0$ .
Not rounding to the stated accuracy; here 3 significant figures.

Things to Be Careful About

The question asks for average speed over the whole motion, not average velocity. The denominator is the total time $37$ s. If a different value of the acceleration time was obtained in part (a), the total time and the trapezium formula should be adjusted accordingly.

Techniques used

compute the durations of each stagefind the area under the velocity-time graphcalculate the total distancedivide total distance by total time

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