A car travels with a uniform velocity of 25 m s^-1 for 5 s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10 s.

Find —

(i) the distance which the car travels before the brakes are applied,

(ii) the retardation and

(iii) the distance travelled by the car after applying the brakes.

(i) As we know,

Distance = Speed x time

Initial velocity = u = 25 m s^-1

Final velocity = 0

time = 5 s

Substituting the values in the formula above we get,

$\text{Distance} = 25 \times 5 \\[0.5em] \Rightarrow \text{Distance} = 125 m$

Hence, distance covered = 125 m.

(ii) Retardation = -a = $\dfrac{v - u}{t}$

t = 10 s

Substituting the values in the formula above we get,

$\text {Retardation} = -\dfrac{0 - 25}{10} \\[0.5em] \Rightarrow \text {Retardation} = \dfrac{25}{10} \\[0.5em] \Rightarrow \text {Retardation} = 2.5 \text {m s}^{-2} \\[0.5em]$

Hence, retardation of the car = 2.5 m s^-2

(iii) As we know,

v² - u² = 2aS

Substituting the values in the formula above we get,

$(0)^2 - (25)^2 = 2 \times (-2.5) \times S \\[0.5em] 0 - 625 = - 5S \\[0.5em] \Rightarrow - 5S = - 625\ \\[0.5em] \Rightarrow S = \dfrac{625}{5} \ \\[0.5em] \Rightarrow S = 125 m \ \\[0.5em]$

Hence, the distance travelled by the car after applying the brakes = 125 m.