Question

11(c) A body starts from rest and travels distances of 120m, 300m and 180m in successive equal time intervals of 12 s. During each interval the body is uniformly accelerated. (i) Calculate the velocity of the body at the end of each successive time interval. (ii) Sketch a velocity-time graph for the motion.

Accepted Answer

For the first motion: distance covered=120m, time interval=12s, initial velocity=0. Using s=ut+½at²: 120=½xax12²; a=(120x2)/144=1.67ms⁻². Using v=u+at: v=(1.67x12)=20m/s.
For the second motion, the final velocity of the first motion becomes the initial velocity of the second (20m/s); distance covered=300m, time interval=12s. Using s=ut+½at²: 300=20(12)+½xax12²; 300-240=½xax144; a=(60x2)/144=0.83ms⁻². v=u+at=20+(0.83x12)=20+10=30m/s.
Also, for the third motion, initial velocity=final velocity of the second motion=30m/s; distance covered=180cm... [as given, 180m], time interval=12s. s=ut+½at²: 180-30(12)=½a(144); 180-360=½a(144); a=(-180x2)/144=-2.5ms⁻² (deceleration). v=u+at=30+(-2.5x12)=30-30=0ms⁻¹, meaning the body decreases to rest at the end of the third interval.

WAEC Physics 2010 Theory — Question 15

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