Question

A bike moves with a constant velocity of 5 m/s for 10 s and then its velocity increases to 10 m/s in next 5 s. thereafterits velocity decreases at a uniform rate until it comes to rest after 10 s. express this entire run of the bike on velocity time graph. from the same graph (a) indentify the time interval when the bike was accelerated (b) calculate the displacement travelled while the bike was deaccelerating ​(c) calculate the average velocity of the bike

Answer 1

Time interval = difference in the time of increase in velocity
= 15s - 10s = 5s

b. distance= area of trapezoid + area of triangle
= 1/2 (AD+OE) x OA + 1/2 BD.CH
= 1/2 (20+25) x 5 + 1/2 10.5 
= 225/2 + 25
= 137.5 m

c. Avg Velocity= S/t
=137.5/ 25
=5.5 m/s