Question

A train starting from a railway station and moving with a uniform acceleration attains a speed of $40\;{\text{km}}/{\text{h}}$ in 10 min, find its acceleration.

Answer 1

The above problem can be solved by using the concept of the kinematics. The acceleration of the train is the same as the change in the speed of the train in certain duration. The train is initially at rest. They accelerate for some time and attain a speed. The change in the speed is equal to the speed after acceleration minus initial speed of the train. The ratio of the change in the speed of the train with time gives the acceleration of the train.

Complete step by step answer
Given
The speed of the train is $v = 40\;{\text{km}}/{\text{h}} = \left( {40\;{\text{km}}/{\text{h}}} \right)\left( {\dfrac{{\dfrac{5}{{18}}\;{\text{m}}/{\text{s}}}}{{1\;{\text{km}}/{\text{h}}}}} \right) = 11.11\;{\text{m}}/{\text{s}}$.
The time taken by the train for acceleration is $t = 10\;\min = 10\;\min \times \dfrac{{60\;{\text{s}}}}{{1\;\min }} = 600\;{\text{s}}$.
The acceleration of the train is given as:
$a = \dfrac{v}{t}......\left( 1 \right)$
Substitute $11.11\;{\text{m}}/{\text{s}}$for v and $600\;{\text{s}}$for t in the equation (1) to find the acceleration of the train.
$a = \dfrac{{11.11\;{\text{m}}/{\text{s}}}}{{600\;{\text{s}}}}$
$a = 0.019\;{\text{m}}/{{\text{s}}^2}$
Thus, the acceleration of the train is $0.019\;{\text{m}}/{{\text{s}}^2}$.

Additional Information The slope of the speed time graph describes the acceleration of the object. The speed of the object is the same as the slope of the distance time graph of the object. The area under the curve of the acceleration time graph describes the speed of the object and the area under the curve of the velocity time graph describes the distance covered by the object.

Note: Be careful in conversion of speed from kilometer per hour to meter per second and of time from minute to seconds. The above problem can also be solved by using the first equation of motion $v = u + at$ and assume the train starts from rest, so substitute u=o in the equation.