Question:
A car moving in a straight line at 30 ms-1 slows uniformly to a speed of 10 ms-1 in 5 sec.Determine the
- (a) Acceleration of the car
- (b) Displacement in the third second
Solution:
(a) Acceleration of the car
Initial velocity of the car u=30 ms-1
Final velocity of the car v=10 ms-1
Time taken t = 5 sec
$${Acceleration =\frac{change \ in \ velocity}{time \ taken}}$$
$${Acceleration=\frac {Final \ velocity -Initial \ velocity} {time \ taken}}$$
$${a=\frac {v-u}{t}}$$
$${a=\frac {10-30}{5}}$$
$${a=\frac{-20}{5}}$$
$${a=-4\frac{m} {s^2}}$$
(b) Displacement in the third second:
$${s_n =u+\frac{a}{2}(2n-1)}$$
$${s_3 =30+\frac{(-4)}{2}[(2(3)-1]}$$
$${s_3=20 \ m}$$
Displacement in the third second is 20 m