A driver applies the brakes on seeing the red | vedclass

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(D) Initial velocity $u = 15 \ m/s$,retardation $a = -0.3 \ m/s^2$,and final velocity $v = 0 \ m/s$ (when it stops).First,calculate the time taken to

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$25$

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(D) Initial velocity $u = 15 \ m/s$,retardation $a = -0.3 \ m/s^2$,and final velocity $v = 0 \ m/s$ (when it stops).First,calculate the time taken to stop: $t = \frac{v-u}{a} = \frac{0-15}{-0.3} = 50 \ s$.Since the vehicle stops in $50 \ s$,which is less than $60 \ s$ (one minute),the displacement after $60 \ s$ is the same as the displacement after $50 \ s$.Using the equation of motion $s = ut + \frac{1}{2}at^2$:$s = (15 	imes 50) + \frac{1}{2} 	imes (-0.3) 	imes (50)^2$$s = 750 - 0.15 	imes 2500 = 750 - 375 = 375 \ m$.The initial distance from the traffic light was $400 \ m$.Therefore,the distance from the traffic light after one minute is $400 \ m - 375 \ m = 25 \ m$.

$A$ driver applies the brakes on seeing the red traffic signal $400 \ m$ ahead. At the time of applying brakes,the vehicle was moving with $15 \ m/s$ and retarding at $0.3 \ m/s^2$. The distance of the vehicle from the traffic light one minute after the application of brakes is: (in $m$)

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