Circular motion -- Car on a circular track

  • Thread starter Nikstykal
  • Start date
  • #1
31
1

Homework Statement


A car has an initial speed v0= 20m/s. It increases its speed along the circular track at s=0, at=(0.9s) m/(s2), where s is in meters. Determine the time needed for the car to travel 35m. The radius of curvature of the track is 40m.

Homework Equations


an = v^2/p
a = dv/dt


The Attempt at a Solution


I really don't know how to approach the problem. I know that I want to define an integral that evaluates a change in s on one side with a change in t on the other.

I started by setting a = dv/dt = dv/ds * ds/dt = v * dv/ds so a * ds = v*dv. By doing this integral from 0 to 35m I found the final velocity to be 26.5 m/s. I do not know where to proceed from here.

Or can i solve that integral without inputting a value for s to get vf = sqrt ( 0.9s^2+400 ) ? Then using that I can do v = ds/dt so dt = (1/v) ds --> t = Integral [ 1/sqrt(0.9s^2+400) ds?
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,537
6,434
Strange question... if I understand it correctly the curvature is irrelevant. Is this perhaps just one part of a longer question?
I get a larger value for the final speed. Please post your working.
The approach you describe at the end looks right, but make it easy on yourself by using symbols for the acceleration function etc., not plugging in numbers: ##\int (ks^2+v_0^2)^{-\frac 12}.ds##.
Can you think of a substitution for s in the integral that will get rid of the square root?
 
  • #3
31
1
Strange question... if I understand it correctly the curvature is irrelevant. Is this perhaps just one part of a longer question?
I get a larger value for the final speed. Please post your working.
The approach you describe at the end looks right, but make it easy on yourself by using symbols for the acceleration function etc., not plugging in numbers: ##\int (ks^2+v_0^2)^{-\frac 12}.ds##.
Can you think of a substitution for s in the integral that will get rid of the square root?
It isn't part of a longer problem, might just be irrelevant information. I know you can integrate that using a trig sub with tan(theta) but it looks a little messy. I was hoping there was another approach.
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,537
6,434
It isn't part of a longer problem, might just be irrelevant information. I know you can integrate that using a trig sub with tan(theta) but it looks a little messy. I was hoping there was another approach.
Tan or sinh. If the integral is correct (and I think it is) then there will not be an easier way.
 
  • #5
31
1
Tan or sinh. If the integral is correct (and I think it is) then there will not be an easier way.
can you explain the sinh substitution method.
 
  • #6
255
30
Being lazy, I'd write it as a quadratic equation: (a/2 X t^2) + 20t - 35 =0. Solve that and choose the value of t that solves the problem.
 
  • #7
31
1
Being lazy, I'd write it as a quadratic equation: (a/2 X t^2) + 20t - 35 =0. Solve that and choose the value of t that solves the problem.
I thought the quadratic form could only be used when you are assuming constant acceleration, where in this case its not, its a(s). In fact, when I use a(s) = a(35) and solve for the values of t, neither answers are correct.
 
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,537
6,434
can you explain the sinh substitution method.
Just as tan2+1=sec2, sinh2+1=cosh2.

I thought the quadratic form could only be used when you are assuming constant acceleration,
Quite so.
 
  • #9
255
30
The quadratic equation should be 0.45t^2 + 20T -35 = 0. Solve using the quadratic formula. I get 1.686 seconds.

By habit I use the simplest method I can. I've been doing this since before pocket calculators, so the simpler the better.
 
  • #10
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,537
6,434
The quadratic equation should be 0.45t^2 + 20T -35 = 0. Solve using the quadratic formula. I get 1.686 seconds.

By habit I use the simplest method I can. I've been doing this since before pocket calculators, so the simpler the better.
I find a better strategy is to use the simplest valid method. As Nikstykal pointed out in post #7, your quadratic method is invalid here. The acceleration is not constant.
 
  • #11
255
30
Being lazy I read the acceleration as 0.9 m/s^2. Oops!
 

Related Threads on Circular motion -- Car on a circular track

  • Last Post
Replies
1
Views
20K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
12
Views
2K
Replies
2
Views
120
Replies
2
Views
6K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
2K
Top