a bus moving at speed of 25m/s beginning to slow at rate 4m/s^2each second how long does it take to stop

Explanation:

Initial velocity is 25m/s i.e. u. Final velocity would be zero since the bus stopped. It is noted that the bus deccelartion of 4m/s, so the acceleration should be considered negative of 4m/s I.e. 4m/s. So using third law of motion i.e.

v^2 - u^2= 2as

(0)^2 - (25)^2= -2×4×s {Note here I used minus sign in acceleration}

-625 = -8×s

s=625 ÷ 8

s= 125 m

So bus would cover a distance of 125m before coming to rest.

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