2024 A/L Physics Past Paper · Question 07 (English Medium)

Answer: (3) Graph (3)

What the question asks

What we know:

The velocity-time graph has four equal parts between the dashed lines.

Part 1: $v$ rises from $0$ in a straight line.

Part 2: $v$ stays the same.

Part 3: $v$ falls in a straight line to $0$.

Part 4: $v$ keeps falling, below $0$.

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

What each letter means:

$s$ ($\text{m}$): the displacement of the object.

$v$ ($\text{m s}^{-1}$): the velocity of the object.

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

The first quote says that the gradient of a displacement-time graph is the velocity of the body. The second quote says that on a curved graph the velocity at a moment is the gradient of the tangent drawn at that point. So we read the velocity graph as the steepness of the displacement graph.

For example, a car pulls away from the lights, speeds up, cruises, then brakes to a stop and backs up a little.

Its distance from the lights grows slowly at first, then steadily, then more and more slowly, stops growing, and finally shrinks a little.

Required equations

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

Reasoning and calculation

Part 1: $v$ grows from $0$

$bends upwards$

The gradient of $s$ starts at $0$ and grows, so the curve starts flat and bends upwards.

Part 2: $v$ stays the same

$straight line$

The gradient stays the same and above $0$, so $s$ rises in a straight line.

Part 3: $v$ falls to $0$

$top at the third line$

The gradient gets smaller and reaches $0$ at the third dashed line. The curve bends over and reaches its top there.

Part 4: $v$ goes below $0$

$falls a little$

The gradient is negative, so $s$ falls. The fall in part 4 is the same size as the rise in part 3, so $s$ comes down a little, not back to $0$.

Check the five graphs.

Graph (3) starts flat and bends up, then is straight, then bends over to a top at the third dashed line, then falls a little. It matches all four parts.

Graph (1) has a sharp peak, graph (2) is flat in the middle, graph (4) is steepest at the start, and graph (5) falls back to zero.

Checks

Check the units. The gradient of the $s$ - $t$ graph is $\text{m} \div \text{s}$, which is $\text{m s}^{-1}$, the unit of velocity.

Does it make sense?

The top of the $s$ curve is where the object stops and turns back. That is where $v = 0$, at the third dashed line. Graph (3) has its top exactly there.

Conclusion

Graph (3) bends up, runs straight, bends over to its top where $v = 0$, then falls a little.

The answer is (3).

Read the velocity graph as the steepness of the $s$ curve: $v$ growing means bending up, $v$ steady means straight, $v = 0$ means a top, $v$ negative means falling.

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