Answer in Molecular Physics | Thermodynamics for Okunola #145052

During the first time interval the body gains the velocity “v(30) = v_0 + a_1cdot 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_2cdot 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}.”