Question

The same retarding force is applied to stop a train. The train stops after 80 m. If the speed is doubled, then what happens to the distance?
A. Same
B. Doubled
C. Halved
D. Four times

Answer 1

Hint: Before we start addressing the problem, we need to know about the retarding force. The force that resists the relative motion of an object is known as retarding force.

Formula Used:
To find the displacement of an object we have,
\[S = \dfrac{{\left( {{v^2} - {u^2}} \right)}}{{2a}}\]
Where,
u is the initial velocity.
v is the final velocity.
a is acceleration.

Complete step by step solution:
They have given that if a force is applied to stop the train and it stops after 80m and when the speed is doubled what happens to the distance. We know the formula to find the displacement is given by,
\[S = \dfrac{{ - \left( {{v^2} - {u^2}} \right)}}{{2a}}\]…….. (1)
Since it is a retarding force, we consider the negative sign in the above equation. Since the force is constant, the acceleration also remains constant. Consider the first case, where they have said that the initial velocity is u and the final velocity, v is zero (because after a certain time it comes to rest).

Now the equation (1) will become,
\[{S_1} = \dfrac{{ - \left( {0 - {u^2}} \right)}}{{2a}}\]
\[ \Rightarrow S = \dfrac{{{u^2}}}{{2a}}\]
Now considering the second case, the speed u will get doubled and v will be zero then, we have to find the distance. From equation (1),
\[{S_2} = \dfrac{{ - \left( {0 - {{\left( {2u} \right)}^2}} \right)}}{{2a}}\]
\[ \Rightarrow {S_2} = \dfrac{{4{u^2}}}{{2a}}\]
\[ \therefore {S_2} = 4{S_1}\]
Therefore, the distance becomes four times when the speed is doubled.

Hence, option D is the correct answer.

Note:The retarding force causes the acceleration of an object to be negative. For example, the velocity of a vehicle decreases when a brake is applied.