A passenger is standing 10 $m$ away from a bus. The bus begins to move with constant acceleration of 5 $m / s^{2}$ away from passenger. With what minimum speed must passenger run to catch bus.

m / s

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m / s

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m / s

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5 \sqrt{2}

m / s

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Solution

The correct option is A 10 $m / s$

Let man catches bus after time $t$ , and let the bus travels a distance $d$ and achieves a speed $v$ .
Then $v = 0 + 5 t$ [For bus]
$v = \frac{10 + d}{t}$ [For man] and $d = \frac{1}{2} \times 5 t^{2}$ [For bus]
Solving all we get is $d = 10 m$ & $v = 10 m / s$ & $t = 2 s$

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A passenger is standing 10 m away from a bus. The bus begins to move with constant acceleration of 5 m/s2 away from passenger. With what minimum speed must passenger run to catch bus.

The correct option is A 10 m/s Let man catches bus after time t, and let the bus travels a distance d and achieves a speed v. Then v=0+5t [For bus] v=10+dt [For man] and d=12×5t2 [For bus] Solving all we get is d=10 m & v=10 m/s & t=2 s

A passenger is standing 10 $m$ away from a bus. The bus begins to move with constant acceleration of 5 $m / s^{2}$ away from passenger. With what minimum speed must passenger run to catch bus.