Mathematics 9709/43 — May/June 2019

Walkthrough

The velocity-time graph is the key to this problem because the distance travelled equals the area under the graph. The motion splits into three phases.

Phase 1 (0 to 5 s): the bus accelerates from rest at $2.1\text{ m s}^{-2}$. On the graph this is a straight line rising from the origin. To find its top speed we use $v = u + at$ with $u = 0$, $a = 2.1$ and $t = 5$:

$$v_{\max} = 2.1 \times 5 = 10.5\text{ m s}^{-1}$$

So the first line segment ends at the point $(5, 10.5)$.

Phase 2 (5 to 29 s): the bus travels at constant speed $10.5\text{ m s}^{-1}$ for 24 s. This is a horizontal line from $t = 5$ to $t = 5 + 24 = 29$ at height 10.5.

Phase 3 (29 to 35 s): the bus decelerates to rest in 6 s. This is a straight line falling from $(29, 10.5)$ to $(35, 0)$. The total time is $5 + 24 + 6 = 35\text{ s}$.

The four corners of the trapezium are $(0,0)$, $(5,10.5)$, $(29,10.5)$ and $(35,0)$.

The area of a trapezium is half the sum of the parallel sides times the height. The parallel sides here are the top edge (24 s long) and the bottom edge (35 s long), and the height is $10.5\text{ m s}^{-1}$:

$$\text{Distance} = \frac{1}{2}(24 + 35)(10.5) = 309.75\text{ m}$$

Alternatively, split the trapezium into two triangles and a rectangle:

$$\frac{1}{2} \times 5 \times 10.5 + 24 \times 10.5 + \frac{1}{2} \times 6 \times 10.5 = 26.25 + 252 + 31.5 = 309.75\text{ m}$$

Key Takeaways

• The area under a velocity-time graph gives the distance travelled.
• A velocity-time graph for motion with constant acceleration consists of straight-line segments.
• The suvat equation $v = u + at$ connects initial speed, acceleration and time.

Common Mistakes

• Not starting the graph at $(0,0)$ — the bus starts from rest, so the first segment must begin at the origin.
• Not ending the graph at $(35,0)$ — the bus stops, so the final velocity is zero.
• Using the wrong parallel sides in the trapezium formula: the parallel sides are the top and bottom edges (24 s and 35 s), not the slanted sides.
• Forgetting that the constant-speed phase runs from $t = 5$ to $t = 29$, so the deceleration begins at $t = 29$.

Things to Be Careful About

• The mark scheme requires the sketch to include the key time values 0, 5, 29 and 35, and the point $(0,0)$.
• Units: velocity in m/s and time in s, so the area has units $\text{m s}^{-1} \times \text{s} = \text{m}$.
• The answer may be given as 309.75 m or rounded to 310 m; both are accepted.