Mathematics 9709/42 — May/June 2020

Walkthrough

Start by identifying the three stages of the motion. The tram is at rest, so the graph starts at the origin. It accelerates uniformly for $20\text{ s}$, so the first section is a straight line rising to the point $(20, V)$. It then travels at constant speed for $170\text{ s}$, so the middle section is horizontal from $t = 20$ to $t = 190$. Finally it decelerates to rest. Because the deceleration magnitude is twice the acceleration, the velocity drops twice as quickly as it rose, so the stopping time is half of $20\text{ s}$, i.e. $10\text{ s}$. The graph therefore ends at $t = 200$. The total time is $20 + 170 + 10 = 200\text{ s}$.

The deceleration slope must be steeper than the acceleration slope because the magnitude of deceleration is larger.

Key Takeaways

A velocity-time graph uses the gradient for acceleration and the area under the graph for distance. The shape of the graph is determined by the sequence of motions: linear increase for uniform acceleration, horizontal line for constant speed, linear decrease for uniform deceleration. Comparing acceleration and deceleration magnitudes tells you how the slopes compare.

Common Mistakes

A common mistake is to forget the deceleration time and write the total time as $20 + 170 = 190\text{ s}$. Another is to draw the deceleration slope with the same steepness as the acceleration slope, although the deceleration magnitude is twice as large. The mark scheme requires a trapezium with the deceleration steeper than the acceleration.

Things to Be Careful About

The total time is not the time at constant speed. The graph must start at the origin because the tram starts from rest, and it must end on the time axis because the tram is brought to rest. The deceleration time must be found from the relationship between acceleration and deceleration, not assumed to be $20\text{ s}$.

Walkthrough

On a velocity-time graph, the distance travelled is the area between the graph and the time axis. The graph from part (a) is a trapezium. The two parallel sides are horizontal: the constant-speed section has length $170\text{ s}$, and the whole motion lasts $200\text{ s}$. The vertical distance between these parallel sides is the constant speed $V$. Use the trapezium area formula $\frac{1}{2}(a+b)h$ with $a = 170$, $b = 200$ and $h = V$. Convert $2.775\text{ km}$ to $2775\text{ m}$ before substituting. Solving $\frac{1}{2}(170+200)V = 2775$ gives $V = 15\text{ m s}^{-1}$.

Key Takeaways

The area under a velocity-time graph gives displacement (distance here because the motion is in one direction). The trapezium area formula is a quick way to combine the three separate areas (acceleration triangle, constant-speed rectangle, deceleration triangle) into one expression.

Common Mistakes

Forgetting to convert kilometres to metres is a common error. Using the wrong parallel sides in the trapezium formula, such as $20$ and $170$, gives the wrong area. The mark scheme expects the equation $\frac{1}{2}(170 + 200)v = 2775$.

Things to Be Careful About

The total time is $200\text{ s}$, not $190\text{ s}$. The constant-speed interval is $170\text{ s}$. The area formula uses the total time and the constant-speed time as the two parallel sides because the trapezium is oriented with its parallel sides horizontal on the time axis. Keep all distances in metres and all times in seconds.

Walkthrough

Acceleration is the rate of change of velocity, so on the velocity-time graph it is the gradient of the first straight section. The tram starts from rest, so initial velocity $u = 0$. After $20\text{ s}$ its velocity is $V$, so using $v = u + at$ gives $V = 20a$. From part (b), $V = 15$, so $15 = 20a$ and $a = 0.75\text{ m s}^{-2}$. The deceleration is twice this, so its magnitude would be $1.5\text{ m s}^{-2}$, but the question asks for the acceleration, not the deceleration.

Key Takeaways

The gradient of a velocity-time graph is acceleration. The constant acceleration formula $v = u + at$ is the algebraic version of this idea. Once one quantity such as $V$ is known, it can be substituted back into an earlier equation to find another quantity.

Common Mistakes

A common mistake is to divide by $10$ (the deceleration time) instead of $20$, or to give the deceleration $1.5\text{ m s}^{-2}$ instead of the acceleration $0.75\text{ m s}^{-2}$. Another is to forget that the tram starts from rest, so $u = 0$.

Things to Be Careful About

The question asks for the magnitude of the acceleration, not the deceleration. Use the acceleration phase time of $20\text{ s}$, not the total time or the deceleration time. The units are $\text{m s}^{-2}$.