Figure shows the velocity-time graph of a particle moving in a straight line.

(i) State the nature of motion of particle.

(ii) Find the displacement of particle at t=6 s.

(iii) Does the particle change its direction of motion?

(iv) Compare the distance travelled by the particle from 0 to 4 s and from 4 s to 6 s.

(v) Find the acceleration from 0 to 4 s and retardation from 4 s to 6 s.

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Solution

(i) From 0 to 4 seconds, the motion is uniformly accelerated and from 4 to 6 seconds, the motion is uniformly retarded.

(ii) Displacement of the particle at 6 s = $\frac{1}{2} \times 6 \times 2$ = 6 m

(iii) The particle does not change its direction of motion as the velocity is always positive.

(iv) Distance travelled by the particle from 0 to 4s (D1) = $\frac{1}{2} \times 4 \times 2$ = 4 m
Distance travelled by the particle from 4 to 6s (D2) = $\frac{1}{2} \times 2 \times 2$ = 2 m
D1:D2 = 4:2 =2:1

(v) Acceleration from 0 to 4 s = (2/4) ms^-2 or 0.5 ms^-2
Retardation from 4 s to 6 s = (2/2) ms^-2 or 1 ms^-2.

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