Engine of a vehicle can give it an acceleration of $1 m s^{- 2}$ and its brakes can retard it at $3 m s^{- 1}$ . The minimum time in which the vehicle can make a journey between stations $A$ and $B$ having a distance of $1200 m$ is

55.6 s

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65.6 s

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50.6 s

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56.5 s

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Solution

The correct option is B $56.5 s$
Let $x$ be the distance traveled until maximum velocity (V) and $y$ be the remaining distance
we can write
$v^{2} - 0 = 2.1. x \Rightarrow x = \frac{v^{2}}{2}$
Similarly
$0 - v^{2} = - 6 y \Rightarrow y = \frac{v^{2}}{6} x + y = 1200 = \frac{{2 v}^{2}}{3} \Rightarrow v = 30 \sqrt{2}$
Now, using equation $v = u + a t$
we get
$t_{1} = 30 \sqrt{2} & t_{2} = 10 \sqrt{2} t_{1} + t_{2} = 40 \sqrt{2} = 56.5$ s

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Engine of a vehicle can give it an acceleration of 1ms−2 and its brakes can retard it at 3ms−1. The minimum time in which the vehicle can make a journey between stations A and B having a distance of 1200m is

The correct option is B 56.5sLet x be the distance traveled until maximum velocity (V) and y be the remaining distancewe can write v2−0=2.1.x⇒x=v22 Similarly0−v2=−6y⇒y=v26x+y=1200=2v23⇒v=30√2 Now, using equation v=u+atwe get t1=30√2&t2=10√2t1+t2=40√2=56.5 s

Engine of a vehicle can give it an acceleration of $1 m s^{- 2}$ and its brakes can retard it at $3 m s^{- 1}$ . The minimum time in which the vehicle can make a journey between stations $A$ and $B$ having a distance of $1200 m$ is