Answer to Question #83329 in Mechanics | Relativity for Raymond

Question

A car runs at a constant speed of 15m/s for 300s and then accelerates uniformly to a speed of 25m/s over a period of 20s. This speed is maintained for 300s before the car is brouglit to rest with uniform retardation in 30s. Calculate the acceleration while the the velocity changes from 15m/s to 25m/s, the total distance covered, and the average speed of the car.

Answer 1

Calculate the acceleration from 15 to 25 m/s in 20 s:

a_2=(v_2-v_1)/t_2 =(25-15)/20=0.5 m/s^2.

The total distance covered:

s=v_1 (t_1+t_2 )+(a_1 t_2^2)/2+v_2 t_3+(v_2 t_4)/2=

=15(300+20)+(0.5·〖20〗^2)/2+25·300+(25·30)/2=12775 m.

<v> =s/(t_1+t_2+t_3+t_4 )=12775/(300+20+300+30)=19.65 m/s.