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A passenger is standing d metres away fr...

A passenger is standing `d` metres away from a bus. The bus begins to move eith constat acceleration `a. To catch the bus the passenger runs at a constant speed (v) towards the bus, at what minimum speed he must have ,so that he may catch the bus.

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Let the passenger catch the bus after time `t`. From 2nd equation of motion, the distance travelled by the bus,
`s_(1) = 0 + (1)/(2) a t^(2)`....(i)
and the distance travelled by the passenger
`s_(2) = ut + 0`....(ii)
Now the passenger will catch the bus if
`d + s_(1) = s_(2)`....(iii)
Substituting the values of `s_(1)` and `s_(2)` from Eqns. (i) and (ii) in (iii), we get
`d + (1)/(2) at^(2) = ut`
i.e., `(1)/(2) at^(2) - ut + d = 0`
or `t = ([u +- sqrt(u^(2) - 2ad)])/(a)`
So, the passenger will catch the bus if t is real, i.e.,
`u^(2) ge 2 ad " or " u ge sqrt(2 ad)`
So, the minimum speed of passenger for catching the bus is `sqrt(2ad)`.
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