Answer to Question #205511 in Physics for Festus

Question

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

Answer 1

The distance covered at the first part of the motion:

$$d_1 = 20m/s\cdot 100s = 2000m$$

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

$$a_2 = \dfrac{30m/s - 20m/s}{15s} = \dfrac23m/s^2$$

The distance covered at the second part of the motion:

$$d_2 = 20m/s\cdot 15 s + \dfrac{2/3m/s^2\cdot (15s)^2}{2} = 375m$$

The total distance is:

$$d = d_1 + d_2 = 2375m$$

Answer. 2375m.