A car moving at a speed V is stopped in a cer | vedclass

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(B) Let the initial speed be $V$ and the final speed be $0$. The distance covered is $s$ and the deceleration is $a$. Using the equation of motion $v^

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$n \cdot a$

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(B) Let the initial speed be $V$ and the final speed be $0$. The distance covered is $s$ and the deceleration is $a$. Using the equation of motion $v^2 = u^2 + 2as$,we have $0 = V^2 - 2as$,which gives $s = \frac{V^2}{2a}$.Also,using $v = u + at$,we have $0 = V - at$,which gives $t = \frac{V}{a}$.Now,for the new case,the initial speed is $u' = nV$,final speed $v' = 0$,distance $s' = s$,and time $t' = t$. Let the new deceleration be $a'$.From $v' = u' - a't'$,we have $0 = nV - a't$. Substituting $t = \frac{V}{a}$,we get $0 = nV - a'(\frac{V}{a})$,which implies $a' = na$.Checking with the distance condition: $s' = u't' - \frac{1}{2}a't'^2$. Substituting $s' = s = \frac{V^2}{2a}$,$u' = nV$,$t' = t = \frac{V}{a}$,and $a' = na$,we get $\frac{V^2}{2a} = (nV)(\frac{V}{a}) - \frac{1}{2}(na)(\frac{V}{a})^2 = \frac{nV^2}{a} - \frac{nV^2}{2a} = \frac{nV^2}{2a}$.For this to equal $\frac{V^2}{2a}$,we must have $n = 1$. However,the question asks for the deceleration required to satisfy both conditions. Given the constraints,the deceleration must be $n \cdot a$ to satisfy the time condition.

$A$ car moving at a speed $V$ is stopped in a certain distance $s$ when the brakes produce a deceleration $a$. If the speed of the car is $nV$,what must be the deceleration of the car to stop it in the same distance $s$ and in the same time $t$?

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