Question

A car starts from rest with constant acceleration. During the 5th second of its motion, it covers a distance of 36m. What is the acceleration and distance covered by the car during this time?

Answer 1

To solve this problem, we can use the equations of motion for uniformly accelerated linear motion. The equation we'll use is:

$s=ut+\frac{1}{2}a{t}^2$

Where:

• $s$ is the distance covered
• $u$ is the initial velocity (which is zero since the car starts from rest)
• $a$ is the acceleration
• $t$ is the time

Given that the car starts from rest ($u=0$), and during the 5th second ($t=5$) it covers a distance of 36m ($s=36$), we can substitute these values into the equation and solve for acceleration ($a$):

$s=ut+\frac{1}{2}a{t}^2$

36=0+12a(5)236 = 0 + frac{1}{2}a(5)^236=0+21​a(5)<span class="