A bus moves away from rest at a bus stop with an acceleration of 1ms^-2. As the bus stars to move a man who is 4m behind the stop runs with a constant speed after the bus. If he just manages to catch the bus find his speed?

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@Callisto @davian @inkyvoyd

I don't know if it works. |dw:1346211401321:dw| Let s be the distance travelled by bus, distance travelled by the man = s+4 Time for the bus and man travelled is the same For the bus: s = ut + (1/2) at^2, where u = 0ms^-1, a = 1ms^-2 For the man, s + 4 = vt No... it doesn't work. I need one more equation :(

yup..i tried to buy the answer is \[2\sqrt{2} ms^{-1}\]

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