Kinematics of Motion in a Straight Line: Question 5

Worked solution

Step 1: Identify what the gradient of a displacement–time graph represents

On a displacement–time graph, the gradient at any point gives the velocity, since $v = \dfrac{ds}{dt}$. Over a straight-line stage, this gradient is simply the change in displacement divided by the change in time.

Step 2: Find the change in displacement and time for this stage

For $4 \le t \le 10$:

• Displacement changes from $s=20$ m (at $t=4$) to $s=5$ m (at $t=10$), so $\Delta s = 5 - 20 = -15\text{ m}$.
• Time taken: $\Delta t = 10 - 4 = 6\text{ s}$.

Step 3: Compute the velocity

$v = \frac{\Delta s}{\Delta t} = \frac{-15}{6} = -2.5$

So the velocity during this interval is $-2.5\text{ m s}^{-1}$. The negative sign shows the particle is moving back toward $O$ (displacement decreasing), consistent with the graph description.

Step 4: Recompute independently as a check

Working from the two endpoints again: the displacement drops by $20-5=15$ m over the $6$ s from $t=4$ to $t=10$. A drop (not a rise) means the gradient must be negative: $-15\div 6 = -2.5$. This matches Step 3 exactly.

Why the other options are wrong

• B ($2.5\text{ m s}^{-1}$): has the correct size but the wrong sign. It ignores that $s$ is decreasing, so the velocity must be negative.
• C ($-3.75\text{ m s}^{-1}$): comes from dividing $-15$ by $4$ (the duration of the first stage) instead of by $6$ (the duration of this stage).
• D ($-15\text{ m s}^{-1}$): is just the change in displacement, without dividing by the time taken to produce a velocity.

Final answer

$\boxed{v = -2.5\text{ m s}^{-1}} \quad \text{(Option A)}$