Kinematics of Motion in a Straight Line: Question 8

Worked solution

Step 1: Recall what area under a velocity–time graph represents

The area under a velocity–time graph gives displacement, but area above the time-axis (positive velocity) counts as positive displacement, while area below the time-axis (negative velocity) counts as negative displacement. The total displacement is the sum of these signed areas.

Step 2: Find the area of each stage

Stage 1 ($0\le t\le4$): a triangle above the axis, base $=4$ s, height $=8\text{ m s}^{-1}$.

$A_1 = \frac12\times4\times8 = 16$

Since the velocity is positive throughout, this contributes $+16$ m to the displacement.

Stage 2 ($4\le t\le6$): a triangle below the axis, base $=2$ s, height $=6\text{ m s}^{-1}$.

$A_2 = \frac12\times2\times6 = 6$

Since the velocity is negative throughout this stage, this contributes $-6$ m to the displacement.

Step 3: Add the signed areas

$\text{displacement} = 16 + (-6) = 10$

So the total displacement at $t=6$ is $10$ m from the starting point.

Step 4: Recompute independently as a check

Total distance travelled (ignoring direction) is $16+6=22$ m, since the particle first moves $16$ m forward, then $6$ m back. Its displacement is therefore the forward distance minus the backward distance: $16-6=10$ m. This matches Step 3 exactly.

Why the other options are wrong

• B ($22\text{ m}$): this is the total distance travelled (adding both areas as positive), not the displacement.
• C ($16\text{ m}$): this only accounts for Stage 1, ignoring the reversal in Stage 2 entirely.
• D ($-6\text{ m}$): this is only the (negative) area of Stage 2, ignoring Stage 1’s contribution.

Final answer

$\boxed{\text{displacement} = 10\text{ m}} \quad \text{(Option A)}$