A student is running to catch a bus, which is stopped at bus stop. when the student is still 40 m from the bus, it starts to pull away, moving with a constant acceleration 0.17 m/s^2. What minimum speed the student must have to catch up with the bus? For what time and what distance does she have to run in that case?
Ans: 3.69 m/s , 21.7 sec, 80 m

Question

A student is running to catch a bus, which is stopped at bus stop. when the student is still 40 m from the bus, it starts to pull away, moving with a constant acceleration 0.17 m/s^2. What minimum speed the student must have to catch up with the bus? For what time and what distance does she have to run in that case?
Ans: 3.69 m/s , 21.7 sec, 80 m

anonymous · Answer

2as=v^2-u^2 gives v I.e bus speed after 1s
Next...

anonymous · Answer

V=u+at  gives time for which student shudder run in order to catch up with the bus.

Distance = speed*time

ujjwal · Answer

Solve it. If you get the given answer finally.. Inform me. you needn't do the whole thing for me then. just give the basic idea. I will solve it myself. and i am pretty familiar with the ideas you are giving me but they are not of much use here..

Right now tell me, do you get the answer which is required at the end?

anonymous · Answer

Got the answers.
Try and see for yourself

ujjwal · Answer

ok

experimentx · Answer

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