Question

A person willing to catch a bus sees the bus standing at 25 m from him. He starts running towards bus with constant speed of 6.25 ms1 and at the same time bus starts moving away from him with constant acceleration of 0.5 ms2. After what time will he catch the bus? Are you getting two values of time? Why are you getting two values? What is physical significance of two values? For what value of speed of the person will you get single value of time? What will happen if the speed of person is less than this?

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Answer 1

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Answer 2

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Explanation:step by step:

D is 25m, speed is 6.25m/s.

By formula

S = d/t

T is 2500/625

T is 4m/s

1st value

V= u + at

U is 0 as he is starting from rest

V is 6.25m/s

Therefore by formula

6.25/ a is T

6.25/0.5 is T

T is 12.5/s

2nd value

We are getting two values because there is both speed and acceleration is given.

The significance is that both are times value