# Pls tell the solution for this urgent

At time t = 0s, velocity of the bus, u = 0

acceleration of the bus, a = 0.9 m/s

^{2}

Initial distance of the passenger behind the bus, d = 20 m

The velocity of the man is given to be v.

A person can reach the bus, if he can cover this gap, d, before the bus attains the velocity v.

Initially, the relative velocity of the man with respect to the bus, u' = v - u = v - 0 = v.

Relative acceleration of the man with respect to the bus, a' = 0 - a = -0.9 m/s

^{2}.

When the man catch the bus, his relative velocity becomes 0.

So,

${0}^{2}=u{\text{'}}^{2}+2a\text{'}d\phantom{\rule{0ex}{0ex}}0={v}^{2}-2\times 0.9\times 20\phantom{\rule{0ex}{0ex}}v=\sqrt{2\times 0.9\times 20}=2\times 3=6m/s$

Thus,

the velocity of the man must be 6 m/s to catch the bus.

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