Question

A train starting from stationary position and moving with uniform acceleration attains a speed of 36 km/h in 10 minutes. Find the acceleration.

Answer 1

Hint : Uniform acceleration signifies that the acceleration is constant i.e. it does not change with time. All the values need to be calculated in the SI unit. So using the formula for the acceleration we can calculate its value by substituting the values of the velocity and time in the question.

Formula used: In this solution we will be using the following formula;
 $ a = \dfrac{{v - u}}{{{t_2} - {t_1}}} $ , where $ a $ is the acceleration, $ v $ is the final velocity considered, $ u $ is the initial velocity, $ {t_1} $ is time at the initial velocity, and $ {t_2} $ is the time at the final velocity.

Complete step by step answer
When a moving body increases its velocity, it is said to have undergone acceleration. For our moving body in the question, it starts from rest and increases velocity to a certain amount under a particular time, hence it accelerates.
In mechanics, acceleration is fundamentally given by
 $ a = \dfrac{{v - u}}{{{t_2} - {t_1}}} = \dfrac{{v - u}}{{\Delta t}} $ , where $ a $ is the acceleration, $ v $ is the final velocity considered, $ u $ is the initial velocity, $ {t_1} $ is time at the initial velocity, and $ {t_2} $ is the time at the final velocity.
Since, the train was stationary, initial velocity is zero. The final velocity considered is 36 km/h. and change in time is ten minutes.
Before proceeding, we must convert all given values to SI units.
 $ v = 36\dfrac{{km}}{h} \times 1000\dfrac{m}{{km}} \times \dfrac{1}{{3600}}\dfrac{h}{s} = 10m/s $
 $ \Delta t = 10\min \times 60\dfrac{s}{{\min }} = 600s $
Hence, by substituting into formula, the acceleration can be calculated as
 $ a = \dfrac{{10 - 0}}{{600}} = \dfrac{1}{{60}} $
 $ \Rightarrow a = 0.0167m/{s^2} $ .

Note
It is customary to write the formula for acceleration as
 $ a = \dfrac{{v - u}}{t} $ , which can then be rearranged to be,
 $ v = u + at $ which is called the first equation of motion.
For clarity, in this format, the time $ t $ signifies the time elapsed too, just like $ {t_2} - {t_1} = \Delta t $ . But, since only the time difference is considered, $ {t_1} $ can be considered a zero reference point and thus made zero as in $ {t_1} = 0 $ , and then $ {t_2} $ can just be called the time taken $ t $ from that time zero.