Answer to Question #145052 in Molecular Physics | Thermodynamics for Okunola

Question

A body at rest is given an initial acceleration of 8.0m/square for 30sec after which the acceleration is reduced to 50m/s square for next 20sec . The body maintains the speed attained for 60sec after which is brought to rest in20sec.draw the velocity time graph

Answer 1

During the first time interval the body gains the velocity $v(30) = v_0 + a_1\cdot \Delta t = 0 + 8.0\,\mathrm{m/s^2}\cdot30\,\text{s} = 240\,\mathrm{m/s}.$

On the second interval the acceleration is said to be 50m/s square, but it may be a typo because of too large value for the quite large time interval, so we'll assume it to be 5.0m/s squared.

$v(50) = v(30) + a_2\cdot \Delta t = 240\,\mathrm{m/s} + 5.0\,\mathrm{m/s^2}\cdot20\,\text{s} = 340\,\mathrm{m/s}.$ This velocity is maintained for 60 second ($a_3=0$) and in 20 sec the body is brought to rest, so the acceleration on this interval is

$a_4 = \dfrac{0-v(50)}{20\,\text{s}} = -17\,\mathrm{m/s}.$