Question

An express train moving at $30m/s$ reduces its speed to 10 m/s in dist6ance of $240 m$. If the breaking force increases by $12.5\%$ in the beginning find the distance that it travels before coming to rest.
A. $270 m$
B. $240 m$
C. $210 m$
D. $195 m$

Answer 1

Hint:We are given initial velocity and final velocity; we are given the displacement covered. We can use the third equation of motion to find the acceleration. Now the force is increased by 12.5%, thus the new force will be 112.5%. Also, we can make use of Newton’s second law to arrive at the solution then.

Complete step by step answer:
First case: Initial velocity, u= 30 m/s, Final velocity, v= 10 m/s, Displacement, s= 240 m
Using Newton’s third equation of motion, \[{{v}^{2}}-{{u}^{2}}=2as\]
$ {{10}^{2}}-{{30}^{2}}=2\times a\times 240 \\
\Rightarrow -800=480a \\
\therefore a=-\dfrac{5}{3}m/{{s}^{2}} \\$
Now from Second law, F= ma
Since, mass does not change any change in the force will be due to change in the acceleration.
New acceleration, a’= 112.5% of a
$a'=1.125\times -\dfrac{5}{3} \\
\Rightarrow a'=-1.875 \\$
Now again we make use of the third equation of motion and this time final velocity will be zero because the body comes to rest.
$v{{'}^{2}}-u{{'}^{2}}=2a's' \\
\Rightarrow 0-{{30}^{2}}=2\times (-1.875)\times s' \\
\Rightarrow -900=-3.75a' \\
\Rightarrow s'=-\dfrac{900}{3.75} \\
\therefore s'=240m$
Thus, the train will come to rest at the distance of $240m$ if the braking force is $12.5\%$ in the initial.

So, the correct option is B.

Note:Since acceleration is defined as the rate of change of velocity and velocity itself is a vector quantity, so, acceleration is also a vector quantity, so it can assume negative values also. A positive value of acceleration means the speed of the body is increasing while the negative value of acceleration means the velocity of the body is decreasing. Negative acceleration also termed as retardation.