Question

A scooter increases its speed from $ 36km/h $ to $ 72km/h $ in $ 10s $ . What is its acceleration?
(A) $ 7.2m/{s^2} $
(B) $ 1m/{s^2} $
(C) $ 3.6m/{s^2} $
(D) $ 10m/{s^2} $

Answer 1

Hint: We know that acceleration is nothing but the rate of change in velocity. Also, we know that the first law of kinematics states that $ v = u + at $ . Also we know that to convert from $ km/h $ to $ m/s $ we need to multiply the magnitude of the former by $ \dfrac{5}{{18}} $ .

Formulas used: We will be using the formula, that is also called as the Newton’s first equation of kinematics, $ v = u + at $ , where $ v $ is the final velocity acquired by the body, $ u $ is the initial velocity of the body, $ a $ is the acceleration achieved by the body, and $ t $ is the time taken by the body to achieve the acceleration $ a $ .

Complete Step by Step solution
As we know the velocity of a body is the rate of change of displacement, $ v = \dfrac{{ds}}{{dt}} $ . Similarly the acceleration the body achieves is the rate of change of velocity, $ a = \dfrac{{dv}}{{dt}} $ . Here $ dv $ is nothing but change in velocity and $ dt $ is nothing but change in time when the body was recorded to be at those velocities.
 $ \Rightarrow dv = v - u $ where $ u $ is the initial velocity and $ v $ is the final velocity.
 $ dt = {t_2} - {t_1} $ where $ {t_2} $ is the time at which the body acquires velocity $ v $ and $ {t_1} $ is the time at which the body acquires velocity $ u $ .
Now we know that initial velocity of the body, $ u = 36km/h $ and the final velocity of the body, $ v = 72km/h $ . We also know that the time required to acquire this change velocity, $ dv $ is $ dt = {t_2} - {t_1} = 10s $ .
We are required to find the value of acceleration, $ a $ . We can also see that the initial and final velocities have different units and we need to convert them from $ km/h $ to $ m/s $ .
We know that $ 1km = 1000m $ , also $ 1hr = 60 \times 60 = 3600s $
 $ \Rightarrow 1km/h = \dfrac{{1000}}{{3600}}m/s = \dfrac{5}{{18}}m/s $
So, the initial velocity, $ u = 36km/hr $ will be $ u = (36 \times \dfrac{5}{{18}})m/s $ .
 $ \Rightarrow u = 10m/s $ .
Similarly, the final velocity, $ v = 72km/h $ will be $ v = (72 \times \dfrac{5}{{18}})m/s $ .
 $ \Rightarrow v = 20m/s $ .
Now that we have the values of $ v,u $ and $ t $ .Let us substitute them in the formula $ v = u + at $ .
Substituting, $ v = 20m/s $ , $ u = 10m/s $ and $ t = 10s $ we get, $ 20 = 10 + a(10) $ .
Subtracting $ 10 $ on both sides,
 $ 10 = 10(a) $
Now by dividing both L.H.S and R.H.S by $ 10 $ we get,
 $ \Rightarrow a = 1m/{s^2} $ .
Hence the correct answer will be option B.

Note:
Alternate method
Once we find the value of $ u = 10m/s $ and $ v = 20m/s $ . We know that $ a = \dfrac{{dv}}{{dt}} $ .
So, $ a = \dfrac{{v - u}}{{{t_2} - {t_1}}} $
 $ \Rightarrow \dfrac{{20 - 10}}{{10}} = 10m/{s^2} $ .
Thus, the problem can also be solved without using the laws of kinematics.