Question #145052

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

Expert's answer

During the first time interval the body gains the velocity v(30)=v0+a1Δt=0+8.0m/s230s=240m/s.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)+a2Δt=240m/s+5.0m/s220s=340m/s.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 (a3=0a_3=0) and in 20 sec the body is brought to rest, so the acceleration on this interval is

a4=0v(50)20s=17m/s.a_4 = \dfrac{0-v(50)}{20\,\text{s}} = -17\,\mathrm{m/s}.



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