A man is at a distance of 6m from a bus. The bus begins to move with a constant acceleration of 3m/s2. In order to catch the bus, the minimum speed with which the man should run towards the bus is :(a) 2 m/s(b) 4m/s© 6m/s(d) 8m/s

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A man is at a distance of 6m from a bus. The bus begins to move with a constant acceleration of 3m/s². In order to catch the bus, the minimum speed with which the man should run towards the bus is :
(a) 2 m/s
(b) 4m/s
© 6m/s
(d) 8m/s

ananya bhat , 7 Years ago

Grade 11

1 Answers

Khimraj

Last Activity: 7 Years ago

let speed is v

then to catch bus

6 + ½*3*t² = v*t

3t² -2vt + 12 = 0

Now the logic is simple, the passenger will catch the bus after a certain time. So the passenger can catch the bus only when time is a real number. So the quadratic equation must give t as a real number.

For that the discriminant (D) of the equation must be greater than or equal to zero.

4v² – 4*3*12 >=0

for min velocity

v² = 36

v = 6 m/s

Hope it clears. If you Like answer then please approve the answer.

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Mechanics> A man is at a distance of 6m from a bus. ...

A man is at a distance of 6m from a bus. The bus begins to move with a constant acceleration of 3m/s2. In order to catch the bus, the minimum speed with which the man should run towards the bus is :(a) 2 m/s(b) 4m/s© 6m/s(d) 8m/s

ananya bhat , 7 Years ago

Khimraj

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