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A body with an initial velocity of 18 km h^(-1) accelerates uniformly at the rate of 9 cm s^(-2) over a distance of 200 m. Calculate : (i) the acceleration in m s^(-2) (ii) its final velocity in m s^(-1). |
Answer» Solution :(i) ACCELERATION =9 cm `s^(-2) = (9)/(100) m s^(-2)` `=0.09 m s^(-2)` (II) Given initial velocity u = 18 km `h^(-1)` `= (18000m)/(60 xx60 s)= 5 m s^(-1)` Acceleration a = 0.09 `m s^(-2) ` and distance S = 200 m From equation of MOTION `v^(2)= u^(2) + 2A S ` `v^(2) = (5)^(2) +2xx0.09 xx200` or ` v^(2)= 25 + 36 = 61 ` `:.` Final velocity v = `sqrt(61) = 7.81 m.s^(-1)` |
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