Question

A train starting from rest attains a velocity of 72km/h in 5 min. Assuming the acceleration is uniform; Find the acceleration and distance travelled by train for this velocity.

Answer 1

Hint: Use the fact whenever a body starts moving from rest, the final velocity attained by that object is a multiplication of the uniform acceleration and time taken to achieve that final velocity. Use laws of motion to calculate the distance travelled.

Complete step-by-step answer:

Given,

Initial velocity\[ = u = 0\,m/s\] 

Final velocity$ = V = 72\,km/h$ 

Time$ = 5\min $ 

We will convert minutes into hours. 

$\dfrac{5}{{60}}h = \dfrac{1}{{12}}h$ 

We have to find out acceleration: we will denote acceleration by $a$ and distance travelled by $s$.Now applying the equation of motion

$a = \dfrac{{v - u}}{t} $

$\Rightarrow a = \dfrac{{72 - 0}}{{\dfrac{1}{{12}}}} $

$\Rightarrow a = 72 \times 12 $

$\Rightarrow a= 864\,km/{h^2} $ 

Hence the acceleration is $864\,km/{h^2}$.

Now, the distance travelled by train:

$s = ut + \dfrac{1}{2}a{t^2} $

$\Rightarrow s= 0 \times \dfrac{1}{{12}} + \dfrac{1}{2}(864){(\dfrac{1}{{12}})^2} $

$\Rightarrow s = \dfrac{1}{2} \times 864 \times \dfrac{1}{{144}} $

$\Rightarrow s = 3\,km $

Hence, the distance travelled by train is 3km.

Note: Basic knowledge of laws of motion is very important, Also remember the fact the distance and displacement are two different things. Whenever the acceleration is slowing the initial speed that is acceleration is in the opposite direction of the velocity it is called deceleration.