tough physics problem

[yalla22]

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?

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[Kussemek]

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

 

[swifteagle43]

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?

Solve for t and that is your answer

 

[fpr85]

good god, I can already tell I'm going to suck when it comes time to take Physics next semester

 

[psiyung]

good god, I can already tell I'm going to suck when it comes time to take Physics next semester

Dont let it scare you. You just take one step at a time when solving physics problems.

 

[Kussemek]

this problem still gives me problems and in the text it is considered one of the hardest in the unit. fyi

 

[swifteagle43]

Believe me man it gets easier after you read the book. This problem is going to be laughable after the first two weeks of class...or atleast the first semester.