Question

a man is 48m behind a bus which is at rest.. the bus starts accelerating at the rate of 1m/s2, at the same time the man starts running with uniform velocity of 10m/s.what is the minimum time in which the man catches the bus

Read 3 Solution.

FEMNA NAZER 9 year ago is this solution helpfull: 2 6
8
FEMNA NAZER 9 year ago is this solution helpfull: 3 6

10t=48+(1/2) * 1 * t^2

t^2 - 20t + 96 = 0

t^2 - 8t - 12t + 96 = 0

t(t-8) -12(t-8)

(t-8)(t-12)

Minimum time = 8

INDANA HAIMASREE 9 year ago is this solution helpfull: 7 3

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