Solveeit Logo
Login

Question

Question: Assertion: Doubling the initial velocity, the stopping distance of moving objects increases by a fac...

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

Explanation

Solution

Hint: One of Newton's laws of motions is necessary to solve this problem, v2−u2=2as{{v}^{2}}-{{u}^{2}}=2as . Putting in the value of final velocity to be zero here, the distance will become the stopping distance dS.{{d}_{S}}.Therefore, dS=−u22a{{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 dS.{{d}_{S}}. Let’s use the newton’s laws of motion, one specific one:
v2−u2=2as{{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.
∴(0)2−u2=2adS⇒−u2=2adS⇒dS=−u22a\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 u2=2u,{{u}_{2}}=2u, the braking deceleration of the vehicle remains the same ‘a’. The new stopping distance is dS2{{d}_{S2}}
Once the brakes are applied again at the new velocity, the vehicle stops now at the new stopping distance of dS2{{d}_{S2}} .
Using the newton’s laws of motion as above of: v2−u2=2as{{v}^{2}}-{{u}^{2}}=2as and substituting the values of this case into it, we get:
∴(0)2−u22=2adS2⇒−(2u)2=2adS2⇒dS2=−4u22a⇒dS2=4(−u22a)⇒dS2=4dS\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 dS2{{d}_{S2}} is 4 times the value of initial stopping distance ofdS,{{d}_{S}},when the value of initial velocity ‘u’ changed to 2 twice the value (u2=2u)({{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 dS=−u22a,{{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, dS∝u2.{{d}_{S}}\propto {{u}^{2}}.