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