# Circular motion -- Car on a circular track

• Nikstykal
In summary: Good thing I didn't try to solve it!In summary, the conversation involved a problem where a car on a circular track increases its speed with an acceleration function of s=0, at=(0.9s) m/(s2). The task at hand was to determine the time needed for the car to travel 35m, with a radius of curvature of 40m. Various methods were discussed, including using an integral and a quadratic equation, but ultimately the correct method involved using a substitution to solve the integral. The final answer for the time needed was 1.686 seconds.

## 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.

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?

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?

haruspex said:
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.

Nikstykal said:
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.

haruspex said:
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.

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.

OldYat47 said:
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.

Nikstykal said:
can you explain the sinh substitution method.
Just as tan2+1=sec2, sinh2+1=cosh2.

Nikstykal said:
I thought the quadratic form could only be used when you are assuming constant acceleration,
Quite so.

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.

OldYat47 said:
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.

Being lazy I read the acceleration as 0.9 m/s^2. Oops!

## What is circular motion?

Circular motion is a type of motion where an object moves in a circular path around a central point.

## What is the difference between uniform circular motion and non-uniform circular motion?

In uniform circular motion, the speed of the object remains constant throughout the circular path, while in non-uniform circular motion, the speed changes at different points along the path.

## How is centripetal force involved in circular motion?

Centripetal force is the force that keeps an object moving in a circular path. It acts towards the center of the circle and is necessary to maintain the object's velocity and prevent it from flying off in a straight line.

## What is the relationship between the radius of the circular track and the speed of the car?

The speed of the car is directly proportional to the radius of the circular track. This means that as the radius increases, the speed of the car also increases.

## How does the mass of the car affect its circular motion on a track?

The mass of the car does not affect its circular motion on a track, as long as the centripetal force is strong enough to keep the car moving in a circular path. However, a heavier car may require a stronger centripetal force to maintain its speed around the track.