A car is moving with a speed of 150 km/h and | vedclass

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(D) The stopping distance $d$ is given by the formula $d = \frac{v^2}{2a}$,where $v$ is the initial velocity and $a$ is the magnitude of deceleration.

Vedclass · Accepted Answer

$3$

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(D) The stopping distance $d$ is given by the formula $d = \frac{v^2}{2a}$,where $v$ is the initial velocity and $a$ is the magnitude of deceleration.Since the braking acceleration $a$ remains constant,the stopping distance is directly proportional to the square of the initial velocity: $d \propto v^2$.Let the initial speed be $v_1 = 150 \ km/h$ and the initial stopping distance be $d_1 = 27 \ m$.The new speed is $v_2 = \frac{1}{3} v_1$.Therefore,the new stopping distance $d_2$ is given by $d_2 = (\frac{1}{3})^2 	imes d_1 = \frac{1}{9} 	imes 27 \ m = 3 \ m$.

Column-$I$	Column-$II$
$(1)$ Velocity-time relation	$(a)$ $v = v_0 + at$
$(2)$ Velocity-displacement relation	$(b)$ $S = v_0t + \frac{1}{2}at^2$
	$(c)$ $v^2 = v_0^2 + 2as$

$A$ car is moving with a speed of $150 \ km/h$ and after applying the brakes,it travels $27 \ m$ before it stops. If the same car is moving with a speed of one-third of the initial speed,then it will stop after traveling how many meters?

Similar Questions

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Column-$I$ Column-$II$

$(1)$ Velocity-time relation $(a)$ $v = v_0 + at$

$(2)$ Velocity-displacement relation $(b)$ $S = v_0t + \frac{1}{2}at^2$

$(c)$ $v^2 = v_0^2 + 2as$

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