2021 A/L Physics Past Paper · Question 12 (English Medium)

Answer: (3) Graph (3)

What the question asks

We want: the velocity-time graph that matches the given displacement-time graph.

The velocity at each moment is the gradient of the $s$-$t$ graph at that moment.

Reading the options

In every option the axes are $v$ and $t$.

So we must find how the gradient of the $s$-$t$ graph changes with time.

Relevant theory

gradient = velocity of the body

When considering the two points X and Y on the graph, the velocity of the ball at these two instances will be given by the gradients of the tangents drawn to the graph at the point X and Y.

Required equations

$v = \dfrac{\Delta s}{\Delta t}$

Reasoning and calculation

The axes of the given graph are displacement $s$ in $\text{m}$ and time $t$ in $\text{s}$.

The axes of the options are velocity $v$ in $\text{m s}^{-1}$ and time $t$ in $\text{s}$.

The $s$-$t$ curve has three parts.

Part 1. It rises, and it gets flatter and flatter.

Part 2. It is flat.

Part 3. It rises again, and it gets steeper and steeper.

Now the gradient of each part.

Part 1. The gradient is positive, and it falls to zero.

Part 2. The gradient is zero.

Part 3. The gradient is positive, and it grows from zero.

The gradient of an $s$-$t$ graph is the velocity.

So $v$ starts large and falls to $0$. Then $v = 0$. Then $v$ rises from $0$.

The velocity is never negative, because $s$ never falls.

Only graph (3) falls to zero, stays at zero, and then rises.

Graph (3).

Checks

$(s2 - s1) / (t2 - t1)$

Does it make sense?

The object moves forward, then stops for a while, then moves forward again.

Its velocity is zero only while it is stopped.

Conclusion

$\text{graph }(3)$

The velocity falls to zero, stays at zero, then rises.

The answer is (3).

Read the gradient, not the height.

A flat part of an $s$-$t$ graph means zero velocity.

A steeper part means a larger velocity.

Exam tip