Average acceleration from velocity-time graph

Steps:

• Note initial velocity at P (t=0) is 0 m/s since the car starts from rest.
• Identify velocity at Q (t=5 s) from the graph endpoint of PQ.
• Calculate average acceleration as change in velocity over time: (v_Q - 0) / 5 s.
• This equals the gradient (slope) of the straight line connecting P to Q.

Why C is correct:

• Average acceleration is defined as Δv/Δt, which on a v-t graph is the gradient of the chord joining the start and end points.

Why the others are wrong:

• A: Area below PQ represents displacement (∫v dt), not acceleration.
• B: Area of triangle PQS represents a portion of displacement, not acceleration.
• D: Gradient of tangent at Q gives instantaneous acceleration at that point, not average over 5 s.

Final answer: C