Mathematics 9709/41 — October/November 2021

Walkthrough

A velocity-time graph is a good way to visualise the three stages. During the first $12\text{ s}$ the bus accelerates uniformly from rest, so its speed rises along a straight line from $0$ to $V$. For the next $30\text{ s}$ its speed is constant, so the graph is horizontal. In the final $6\text{ s}$ it decelerates uniformly to rest, so the line falls from $V$ to $0$. The total distance travelled equals the total area under this graph.

Break the area into three parts:

$$\frac{1}{2}(12)(V) = 6V$$

for the first triangle,

$$30V$$

for the rectangle, and

$$\frac{1}{2}(6)(V) = 3V$$

for the final triangle. Their sum is the total distance:

$$6V + 30V + 3V = 585$$

Solving gives $V = 15$, so the constant speed is $15\text{ ms}^{-1}$.

Key Takeaways

• The area under a velocity-time graph gives displacement or total distance.
• Piecewise motion can be split into triangles and rectangles on the velocity-time graph.
• Setting up one equation in the unknown maximum speed is enough to solve the problem.

Common Mistakes

• Forgetting the factor of $\frac{1}{2}$ for the triangular accelerating and decelerating areas.
• Using $12V$ instead of $6V$ for the first phase.
• Not showing a complete method to obtain an equation in $V$, which is required by the mark scheme.
• Giving a negative speed or an unsupported value.

Things to Be Careful About

• The constant speed must be positive; the mark scheme requires the final speed to be positive.
• Make sure the times used match the correct phases: $12\text{ s}$, $30\text{ s}$ and $6\text{ s}$.
• A velocity-time graph is acceptable as direct working; if using suvat equations, you must still form a complete equation in $V$.

Walkthrough

Once $V = 15\text{ ms}^{-1}$ is known, consider only the final $6$ seconds. The initial speed is $u = 15$, the final speed is $v = 0$, and the time is $t = 6$. Using $v = u + at$,

$$0 = 15 + 6a$$

so

$$a = -2.5\text{ ms}^{-2}$$

The negative sign tells us that the acceleration is opposite to the direction of motion. Since the question asks for the magnitude of the deceleration, report the positive value $2.5\text{ ms}^{-2}$.

Key Takeaways

• The suvat equation $v = u + at$ links speed, time and acceleration.
• Deceleration is an acceleration in the opposite direction; its magnitude is a positive number.
• The answer to a 'magnitude' question must be positive.

Common Mistakes

• Giving $a = -2.5\text{ ms}^{-2}$ as the magnitude; the mark scheme explicitly disallows this.
• Using the wrong time interval, e.g. using $12\text{ s}$ or $48\text{ s}$ instead of $6\text{ s}$.
• Misreading 'deceleration' as a negative magnitude.

Things to Be Careful About

• The final answer should be the magnitude, so write $2.5$, not $-2.5$.
• Use consistent units: speed in $\text{ms}^{-1}$, time in seconds, acceleration in $\text{ms}^{-2}$.