Solving 615-kg Racing Car Homework Part 1

[ilkjester]

Homework Statement

1) A 615-kg racing car completes one lap in 14.3 s around a circular track with a radius of 50.0 m.
The car moves at constant speed.
a) what is the acceleration of the car?
b) what force must the track exert of the tires to produce this acceleration.

Homework Equations

Ac=v^2/r
Fnet=mv^2/r

The Attempt at a Solution

ac=14.3^2/50.0
ac=4m/s

Now I am sure I did the equations right but the teacher wanted to know the direction of the acceleration I believe. Because some other kids asked him if there was any acceleration at all because it wasn't slowing down or speeding up.

 

[learningphysics]

You didn't do the calculation right. 14.3 is the time for one lap. It isn't v... What is the distance traveled in 1 lap... divide that distance by 14.3... that gives your v.

The direction of acceleration is towards the center of the circular track... whenever an object is moving at a fixed speed in a circular path, the acceleration is towards the center at any time...

 

[ilkjester]

You didn't do the calculation right. 14.3 is the time for one lap. It isn't v... What is the distance traveled in 1 lap... divide that distance by 14.3... that gives your v.

The direction of acceleration is towards the center of the circular track... whenever an object is moving at a fixed speed in a circular path, the acceleration is towards the center at any time...

Ok I was thinking before that I could be wrong. So what I did was draw a circle and labeled the radius 50.0 m I was then going to try to find the length of an arc of the circle but I can't remember how.

 

[learningphysics]

Ok I was thinking before that I could be wrong. So what I did was draw a circle and labeled the radius 50.0 m I was then going to try to find the length of an arc of the circle but I can't remember how.

circumference of a circle = $$2\pi r$$. so that would be the distance of 1 lap.

 

[ilkjester]

circumference of a circle = $$2\pi r$$. so that would be the distance of 1 lap.

lol i just did that your way after finding the length of the track and its the same.
$$2\pi r$$=314/14.3=22

 

[learningphysics]

yeah, that's right. might want to keep a few extra decimal places and just round at the end... I get v = 21.969m/s

then you can get a = v^2/r using that.

 

[ilkjester]

yeah, that's right. might want to keep a few extra decimal places and just round at the end... I get v = 21.969m/s

then you can get a = v^2/r using that.

Thanks for the help I had a feeling I was doing something wrong.

 

[learningphysics]

Thanks for the help I had a feeling I was doing something wrong.

no prob.

 

[ilkjester]

b) what force must the track exert of the tires to produce this acceleration.
for this i used the fnet=mac
fnet=615kg(9.65)
fnet=5934.75
after i told the other class mates that were saying there was no acceleration at all how you do the problem and that there was indeed an acceleration they asked me what the coefficient of friction was. I don't know if i was suppose to find that because i think the way i found the force was right.

 

[learningphysics]

b) what force must the track exert of the tires to produce this acceleration.
for this i used the fnet=mac
fnet=615kg(9.65)
fnet=5934.75
after i told the other class mates that were saying there was no acceleration at all how you do the problem and that there was indeed an acceleration they asked me what the coefficient of friction was. I don't know if i was suppose to find that because i think the way i found the force was right.

your force looks right to me. but I get 5936.53 after carrying the dec. places...

There is definitely an acceleration. Otherwise the car would go in a straight line... if an object changes directions... that means it accelerates... velocity is speed with a direction... even if speed remains constant... if the direction changes, then the velocity changes... so accleeration happens.

We can't get the coefficient of friction without some extra piece of information... but it isn't asked... but friction is acting, that's for sure.

 

[ilkjester]

There is definitely an acceleration. Otherwise the car would go in a straight line... if an object changes directions... that means it accelerates... velocity is speed with a direction... even if speed remains constant... if the direction changes, then the velocity changes... so accleeration happens.

We can't get the coefficient of friction without some extra piece of information... but it isn't asked... but friction is acting, that's for sure.

thats what i told them except for the straight line thing. So the way i found force was right. Thats how i told them to do it. I am not sure why they thought they had to find to friction. You have been the biggest help ever by the way.

 

[learningphysics]

thats what i told them except for the straight line thing. So the way i found force was right. Thats how i told them to do it. I am not sure why they thought they had to find to friction. You have been the biggest help ever by the way.

you're welcome. glad to help. :)