# Pls tell the solution for this urgent

Solution,
At time t = 0s, velocity of the bus, u = 0
acceleration of the bus, a = 0.9 m/s2
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/s2.
When the man catch the bus, his relative velocity becomes 0.
So,
${0}^{2}=u{\text{'}}^{2}+2a\text{'}d\phantom{\rule{0ex}{0ex}}0={v}^{2}-2×0.9×20\phantom{\rule{0ex}{0ex}}v=\sqrt{2×0.9×20}=2×3=6m/s$
Thus,
the velocity of the man must be 6 m/s to catch the bus.

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