Question

Assertion: Doubling the initial velocity, the stopping distance of moving objects increases by a factor of 4 (for the same deceleration).
Reason: Stopping distance is proportional to the square of the initial velocity.
A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C. Assertion is correct but Reason is incorrect
D. Both Assertion and Reason are incorrect

Answer 1

Hint: One of Newton's laws of motions is necessary to solve this problem, $ {{v}^{2}}-{{u}^{2}}=2as $ . Putting in the value of final velocity to be zero here, the distance will become the stopping distance \[{{d}_{S}}.\]Therefore, $ {{d}_{S}}=\dfrac{-{{u}^{2}}}{2a} $ . Changing the initial velocity for the assertion condition and finding out the stopping distance will give us the solution to this problem.

Complete step-by-step answer:
Let’s consider, a vehicle is travelling at an initial velocity ‘u’. It has a braking acceleration ‘a’. The stopping distance is \[{{d}_{S}}.\] Let’s use the newton’s laws of motion, one specific one:
 $ {{v}^{2}}-{{u}^{2}}=2as $
The final velocity ‘v’ of the vehicle will be zero, since the vehicle will come to a rest after covering the stopping distance from the moment brakes were applied.
 $ \therefore {{(0)}^{2}}-{{u}^{2}}=2a{{d}_{S}}\Rightarrow -{{u}^{2}}=2a{{d}_{S}}\Rightarrow {{d}_{S}}=\dfrac{-{{u}^{2}}}{2a} $
Let’s increase the initial velocity of the vehicle to twice the initial velocity, that is $ {{u}_{2}}=2u, $ the braking deceleration of the vehicle remains the same ‘a’. The new stopping distance is $ {{d}_{S2}} $
Once the brakes are applied again at the new velocity, the vehicle stops now at the new stopping distance of $ {{d}_{S2}} $ .
Using the newton’s laws of motion as above of: $ {{v}^{2}}-{{u}^{2}}=2as $ and substituting the values of this case into it, we get:
 $ \therefore {{(0)}^{2}}-u_{2}^{2}=2a{{d}_{S2}}\Rightarrow -{{(2u)}^{2}}=2a{{d}_{S2}}\Rightarrow {{d}_{S2}}=\dfrac{-4{{u}^{2}}}{2a}\Rightarrow {{d}_{S2}}=4(\dfrac{-{{u}^{2}}}{2a})\Rightarrow {{d}_{S2}}=4{{d}_{S}} $
Therefore, the new stopping distance $ {{d}_{S2}} $ is 4 times the value of initial stopping distance of\[{{d}_{S}},\]when the value of initial velocity ‘u’ changed to 2 twice the value $ ({{u}_{2}}=2u) $ .
Hence the assertion is true. Further, both the values of stopping distance are proportional to square of their respective initial velocities, which causes it. This means, both assertion and reason is true and reason is a correct explanation of assertion.

Note: Since, the reason states that stopping distance is proportional to the square of initial velocity, hence, even though $ {{d}_{S}}=\dfrac{-{{u}^{2}}}{2a}, $ the fact that initial velocity squared is available in RHS is enough to state that the reason is true. The negative sign wouldn’t matter either. That is, $ {{d}_{S}}\propto {{u}^{2}}. $