Motion in One Dimension — Question 13

(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 = $\dfrac{\text v - \text u}{\text t}$

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,

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

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