Question #205511

A car move at a constant velocity of 20m/s for 100seconds and then accelerated uniformly to a velocity of 30m/s over 15 seconds. Calculate the distance covered during the motion


Expert's answer

The distance covered at the first part of the motion:


d1=20m/s100s=2000md_1 = 20m/s\cdot 100s = 2000m

By definition, the acceleration at the second part of the motion:


a2=30m/s20m/s15s=23m/s2a_2 = \dfrac{30m/s - 20m/s}{15s} = \dfrac23m/s^2


The distance covered at the second part of the motion:


d2=20m/s15s+2/3m/s2(15s)22=375md_2 = 20m/s\cdot 15 s + \dfrac{2/3m/s^2\cdot (15s)^2}{2} = 375m

The total distance is:


d=d1+d2=2375md = d_1 + d_2 = 2375m

Answer. 2375m.


LATEST TUTORIALS
APPROVED BY CLIENTS