Question

A body starts from rest in a straight line under a force causing a constant acceleration of 2 m s–2.
After 10s the force is removed and the body comes to rest in 20 seconds. Find the distance travelled by the body in the last 20s.

Answer 1

Answer on Question #53586, Physics Mechanics Kinematics Dynamics

A body starts from rest in a straight line under a force causing a constant acceleration of $2\,\mathrm{m\,s^{-2}}$. After 10s the force is removed and the body comes to rest in 20 seconds. Find the distance travelled by the body in the last 20s.

Solution

After 10s the force is removed the speed of the body is $v_{1} = a_{1} \cdot t_{1}$.

In the last 20s. $v_{1} + a_{2}t_{2} = 0$

The second acceleration is $a_{2} = -v_{1} / t_{2} = -a_{1}t_{1} / t_{2}$

The distance travelled by the body in the last 20s is

$$s = v_{1}t_{2} + \frac{a_{2}t_{2}^{2}}{2} = (a_{1} \cdot t_{1})t_{2} + \frac{(-a_{1}t_{1} / t_{2})t_{2}^{2}}{2} = a_{1}t_{1}t_{2} - a_{1}t_{1}t_{2} / 2 = a_{1}t_{1}t_{2} / 2 = 2(m / s^{2}) \cdot 10s \cdot 20s = 400\,\mathrm{m}$$

Answer: $s = a_{1}t_{1}t_{2} / 2 = 400\,\mathrm{m}$

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