A car is moving in a straight line with speed 18 km h-1. It is stopped in 5s by applying the brakes. Find — (i) the speed of car in m s -1, (ii) the retardation and (iii) the speed of car after 2 s of applying the brakes .

Speed of car = 18 km h-1

(i) To convert speed to m s-1

18 km h-1 = 18\ km\over 1\ h

= 18 \times 1000\ m\over 60 \times 60\ s

= 18 \times 10\ m\over 6 \times 6\ s

= 5 m s<sup>-1</sup>

Hence, 18 km h-1 is equal to 5 m s-1.

(ii) As the car is stopped, the final velocity = 0

Initial velocity (u) = 18 km h-1 = 5 m s-1

Time (t) = 5 s

Acceleration (a) = v-u\over t

= 0-5\over 5

= -1 m s-2

Hence, acceleration of the car is -1 m s-2. Negative sign shows that the velocity decreases with time, so retardation is 1 m s -2.

(iii) Time (t) = 2 s

acceleration (a) = -1 m s-2

Initial velocity (u) = 5 m s-1

Substituting the values in the formula for acceleration, we get,

Acceleration (a) = v-u\over t

or, -1 = v-5\over 2

or, -2 = v – 5

or, v – 5 = -2

or, v = 3 m s-1

Hence, the speed of the car after 2 s of applying the brakes = 3 m s-1.