Non Uniform Circular Motion Problem

[moneenfan]

A new car is tested on a 200m diameter track. If the car speeds up at a steady 1.5 m/s^2, how long after starting is the magnitude of its centripetal acceleration equal to the tangental acceleration?

So we know that our acceleration is equal to 1.5 m/s^2, our radius of our circle is 100m. I am not sure what the formulas i need in order to solve.

Any help?

 

[tiny-tim]

hi moneenfan!

(try using the X² icon just above the Reply box )

A new car is tested on a 200m diameter track. If the car speeds up at a steady 1.5 m/s^2, how long after starting is the magnitude of its centripetal acceleration equal to the tangental acceleration?

the tangental acceleration is 1.5 m/s² …

and i assume you know a formula relating centripetal acceleration to speed?

 

[moneenfan]

well i know that Fnet(r)=ma(r)=mv^2/r=mw^2r
Fnet(t)=0 if its uniform circular motion or ma(t) for non uniform circular motion
Fnet(z)=0

Whats messing me up exactly is that were not given a mass only an acceleration.
Im not sure how to approach this

 

[tiny-tim]

Whats messing me up exactly is that were not given a mass only an acceleration.
Im not sure how to approach this

Call the mass m … it'll cancel out when you do F = ma anyway!

 

[rwisz]

To solve this problem, we can use the formulas for centripetal acceleration and tangential acceleration. Centripetal acceleration is given by the formula a = v^2/r, where v is the velocity and r is the radius of the circle. Tangential acceleration is given by the formula a = dv/dt, where dv is the change in velocity and dt is the change in time.

Since we know the acceleration and radius, we can use the formula for centripetal acceleration to find the velocity of the car at any point on the track. Then, we can use the formula for tangential acceleration to calculate the change in velocity over time.

At the beginning of the track, the car is starting from rest, so the velocity is 0 m/s. We can plug this into the formula for centripetal acceleration to find the initial acceleration as a = 0^2/100 = 0 m/s^2.

As the car speeds up at a steady 1.5 m/s^2, the centripetal acceleration will also increase. At some point, the magnitude of the centripetal acceleration will be equal to the tangential acceleration, which is also 1.5 m/s^2.

To find the time at which this happens, we can set the two formulas equal to each other and solve for t:

1.5 m/s^2 = dv/dt

We know that dv is equal to the change in velocity, which is the final velocity (v) minus the initial velocity (0). So we can rewrite the equation as:

1.5 m/s^2 = (vf - 0)/t

Solving for t, we get t = vf/1.5 m/s^2.

Since we know that the final velocity is equal to the velocity at any point on the track, we can use the formula for centripetal acceleration to find the final velocity:

1.5 m/s^2 = vf^2/100

Solving for vf, we get vf = 10 m/s.

Plugging this value into our equation for t, we get t = 10 m/s / 1.5 m/s^2 = 6.67 seconds.

Therefore, after 6.67 seconds, the magnitude of the centripetal acceleration will be equal to the tangential acceleration.