tough physics problem

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yalla22

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A runner hopes to complete the 10,000 m run in less than 30 min. After exactly 27 min there are still 1100 m to go. The runner must then acceleratie at .2m/s2 for how many seconds in order to achieve the desired time?

is this out of giancoli?

Vnot=deltax/deltat=(10000m-1100m)/(27min)(60s)=5.49m/s
The runner has to cover the ending 1100 in 180s, If he accelerates in that time it will be
v=Vnot+at=5.49+(.20)t
and the distance covered during the acceleration will be
x1=VnotT+.5at^2=5.49t+.5(.20)t^2
Remaining DIstance:
1100m-x1=v(180s-t)

convert to quadratic equation and you get t=3.1s

A runner hopes to complete the 10,000 m run in less than 30 min. After exactly 27 min there are still 1100 m to go. The runner must then acceleratie at .2m/s2 for how many seconds in order to achieve the desired time?

Aka
Vo= ((10,000m-1100m)/(27minuts * 60 seconds))(We assume it is constant velocity) = 5.49 m/s

A= .2 m/s^2

The magic equation you are looking for is
D= Vo*t+(1/2)*a*t^2

D=(1100)

So

1100=5.49*t+(1/2)*(.2)*t^2

right?