The two person of a body undergoing a uniformly accelerated motion are (10s, 10$ms^{-1}$) and (30s, 50$ms^{-1}$). On the velocity-time graph pf the body.

Analyze the question: The question is a bit garbled, but it says two points on the velocity-time graph of a body in uniformly accelerated motion are (10 s, 10 m/s) and (30 s, 50 m/s). The usual WASSCE question asks for the acceleration of the body, which is the slope of the graph.

Concept: On a velocity-time graph, the slope equals the acceleration:

a = (v₂ − v₁) / (t₂ − t₁)

Solution:

• Change in velocity = 50 − 10 = 40 m/s
• Change in time = 30 − 10 = 20 s

a = 40 / 20 = 2.0 m/s²

Check against the options: 2.

Answer: B (2.0 m/s²)

Tip: When you see points on a velocity-time graph, think "slope = acceleration" and "area under the graph = distance." Always subtract in the same order (final minus initial) for both velocity and time so the signs stay consistent.