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Question
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 = (1/2) (6) (2) = 6 m
(iii) The particle does not change its direction of motion.
(iv) Distance travelled by the particle from 0 to 4s (D1) = (1/2) (4) (2) = 4 m
Distance travelled by the particle from 4 to 6s (D2) = (1/2) (2) (2) = 2 m
D1:D2:: 4:2
D1:D2:: 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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