Question

A car starts from rest and accelerated uniformly for 5s until it attains a velocity of 3m^-1. It then travels with uniform velocity for 15s before decelerates uniformly to rest in 10s

1. Calculate the acceleration during the first 5 seconds and deceleration during the last 10 seconds
2. Total distance covered through the motion

Answer 1

1) Let's first calculate the acceleration of the car during the first 5 seconds:

$$v_1=v_{01}+a_1t_1,$$ $$a_1=\dfrac{v_1-v_{01}}{t_1}=\dfrac{3\ \dfrac{m}{s}-0}{5\ s}=0.6\ \dfrac{m}{s^2}.$$

Then, we can find the acceleration of the car during the last 10 seconds:

$$v_3=v_{03}+a_3t_3,$$ $$a_3=\dfrac{v_3-v_{03}}{t_3}=\dfrac{0-3\ \dfrac{m}{s}}{10\ s}=-0.3\ \dfrac{m}{s^2}.$$

The sign minus means that the car decelerates.

2) We can find the total distance covered as follows:

$$d_{tot}=d_1+d_2+d_3,$$ $$d_1=v_{01}t_1+\dfrac{1}{2}a_1t_1^2=\dfrac{1}{2}\cdot0.6\ \dfrac{m}{s^2}\cdot(5\ s)^2=7.5\ m,$$ $$d_2=v_{02}t_2+\dfrac{1}{2}a_2t_2^2=3\ \dfrac{m}{s}\cdot15\ s=45\ m,$$ $$d_3=v_{03}t_3+\dfrac{1}{2}a_3t_3^2=3\ \dfrac{m}{s}\cdot10\ s+\dfrac{1}{2}\cdot(-0.3\ \dfrac{m}{s^2})\cdot(10\ s)^2=15\ m,$$ $$d_{tot}=7.5\ m+45\ m+15\ m=67.5\ m.$$