2022 A/L Physics Past Paper · Question 19 (English Medium)

Answer: (1) Only (A) is true.

What the question asks

We want: which of the statements (A), (B) and (C) are true for the two graphs.

What each letter means:

$s$ ($\text{m}$): the displacement of an object from its starting point.

$t$ ($\text{s}$): the time.

$v$ ($\text{m s}^{-1}$): the velocity of an object. It is the gradient of its $s$–$t$ graph.

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.

That is, the ball has travelled with a decreasing velocity, or with a "deceleration" vertically up wards.

Reading the options

The options are combinations of (A), (B) and (C).

We test each statement on its own.

For example, think of two runners on a straight road. One starts well ahead and tires. The other starts at the line and keeps speeding up.

At some moment the second runner catches up. They are at the same place then. But the second runner is still speeding up, while the first is slowing down. So the two runners are not equally fast at that moment.

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

In words: the velocity is the gradient of the $s$–$t$ graph.

Meeting at a point on the graph means the same $s$. It says nothing about the same gradient.

Reasoning and calculation

Both objects have velocities in the same direction.

Both graphs go up as $t$ increases, so the gradient is positive for both. Both objects move in the same direction.

A

Velocities of both objects increase with time.

The graph of $P$ gets steeper, so its velocity increases. The graph of $Q$ gets flatter, so its velocity decreases. Not both increase.

B

Both objects have the same velocity at the crossing point of the two graphs.

At the crossing both graphs have the same $s$, so the objects are at the same place. But the graph of $P$ is steeper there than the graph of $Q$. So the velocity of $P$ is bigger.

C

Statement (A) is true: both gradients are positive.

Statement (B) is false: the velocity of $Q$ decreases.

Statement (C) is false: at the crossing the gradients are different.

So only (A) is true.

Why the other options fail

Statement (B) is false, because the graph of $Q$ gets flatter.

Statement (C) is false. The objects meet at the crossing, but their gradients are different.

Statements (B) and (C) are both false, because the gradient of $Q$ falls and the gradients differ at the crossing.

Statements (B) and (C) are false, so not all three are true.

Conclusion

Only statement (A) is true. (B) and (C) are false.

The answer is (1).

The velocity is the gradient of an $s$–$t$ graph, not its height.

Two graphs that cross are at the same place. They are not necessarily equally fast.

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