A bus is moving with the initial velocity of $^{'} u^{'} m / s$ . After applying the breaks, its retardation is $0.5 m / s^{2}$ and it stopped after $12 s$ . Find the initial velocity (u) and distance travel by the bus after applying the breaks.

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Solution

Final velocity $v = 0$
Using $v = u + a t$
So, $0 = u + (- 0.5) \times 12$
$⟹ u = 6 m / s$
Using, $v^{2} - u^{2} = 2 a S$
Or, $0 - 6^{2} = 2 \times (- 0.5) \times S$
$⟹ S = 36 m$

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A bus is moving with the initial velocity of ′u′m/s. After applying the breaks, its retardation is 0.5m/s2 and it stopped after 12s. Find the initial velocity (u) and distance travel by the bus after applying the breaks.

Final velocity v=0Using v=u+atSo, 0=u+(−0.5)×12⟹ u=6 m/sUsing, v2−u2=2aSOr, 0−62=2×(−0.5)×S⟹ S=36 m

A bus is moving with the initial velocity of $^{'} u^{'} m / s$ . After applying the breaks, its retardation is $0.5 m / s^{2}$ and it stopped after $12 s$ . Find the initial velocity (u) and distance travel by the bus after applying the breaks.

Final velocity $v = 0$
Using $v = u + a t$
So, $0 = u + (- 0.5) \times 12$
$⟹ u = 6 m / s$
Using, $v^{2} - u^{2} = 2 a S$
Or, $0 - 6^{2} = 2 \times (- 0.5) \times S$
$⟹ S = 36 m$