Question

A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 12m/s.Its velocity decreases at a uniform rate of 0.5m/s^2. [a] Find the time it will take to come to rest. [b] Find the distance covered by it before coming to rest?

Answer 1

u = 12 m/s
a = - 0.50 m/s/s

v = 0
     v = u + a t
  => t = (0 - 12 ) / (-0.50) =  24 sec

  s = u t + 1/2 a t²  = 12 * 24  -  1/2 * 0.50 * 24²
       = 144 meters

Answer 2

Answer:

a) It takes 4 seconds to come to rest.

b) The distance covred before coming to rest is 144 meter.

Explanation:

Initial velocity Vi = 12 m/S

a = deceleration rate =  $-0.5 \mathrm{m} / \mathrm{s}^{2}$

From Laws of Kinematics,

$\begin{array}{l}{\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}+\mathrm{at} \quad \rightarrow(1)} \\ {\mathrm{d}=\mathrm{Vit}+1 / 2 \mathrm{at}^{2} \rightarrow(2)}\end{array}$

$V_{f}$ = Final velocity = 0 as object comes to rest.

Time taken to come to rest, t = $\bold{\left(\mathrm{V}_{\mathrm{f}}-\mathrm{V}_{\mathrm{i}}\right) / \mathrm{a}}$ rest.  = 24 Second

d = distance travelled =  $V_{i t}+1 / 2 a t^{2}=12 * 24+1 / 2(-0.5)\left(24^{2}\right)$

So d = 144 meter