A race car starts from rest on a circular track of radius 500 m. The car's speed increases at the constant rate of 0.520 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the following.

a)speed of the race car (m/s)
b)the distance traveled (m)
c)elapsed time (s)

The answer that is on the other questions asked is wrong...please help!

a(τ)=0.52 m/s², R=500 m

v=? s=? t=?
------
v=a(τ) •t
centripetal (normal) acceleration is
a(n)=v²/R =( a(τ) •t)²/R = a(τ),
a(τ) ² •t²/R = a(τ),
t=sqrt{R/a(τ)}=sqrt{500/0.52)= 31 s,
s=a(τ) •t²/2 = 0.52•31²/2 =249.9 m,
v= a(τ) •t =0.52•31 = 16.12 m/s.

To solve this problem, we need to find the point where the magnitudes of the centripetal and tangential accelerations are equal.

Let's analyze the problem step by step:

Step 1: Find the magnitude of the centripetal acceleration at any point on the circular track.
The centripetal acceleration is given by the formula:
a_c = v^2 / r
where v is the velocity of the car and r is the radius of the circular track.

Step 2: Find the magnitude of the tangential acceleration at any point on the circular track.
The tangential acceleration is given by the formula:
a_t = d(v) / dt
where d(v) is the change in velocity and dt is the change in time.

Step 3: Equate the magnitudes of the centripetal and tangential accelerations and solve for v, the speed of the race car.

Step 4: Use the speed of the race car to find the distance traveled and the elapsed time.

Now let's solve the problem:

Step 1: Find the magnitude of the centripetal acceleration.
a_c = v^2 / r
Plugging in the values given:
0.520 m/s^2 = v^2 / 500 m

Step 2: Find the magnitude of the tangential acceleration.
a_t = d(v) / dt
Since the speed increases at a constant rate, the tangential acceleration is equal to the rate of change of speed:
a_t = 0.520 m/s^2

Step 3: Equate the magnitudes of the centripetal and tangential accelerations.
0.520 m/s^2 = v^2 / 500 m

Simplifying the equation:
0.520 m/s^2 * 500 m = v^2
v^2 = 260 m^2/s^2
v = √(260) m/s
v ≈ 16.12 m/s

So, the speed of the race car is approximately 16.12 m/s (a).

Step 4: Use the speed of the race car to find the distance traveled and the elapsed time.
To find the distance traveled, we need to know the time it took.

We can use the equation of motion:
v = u + at
where u is the initial velocity, a is the acceleration, t is the time, and v is the final velocity.

In this case, the race car starts from rest (u = 0 m/s), so the equation becomes:
v = at

Solving for t:
t = v / a
t = 16.12 m/s / 0.520 m/s^2
t ≈ 31.00 s

To find the distance traveled, we can use the equation:
s = ut + (1/2)at^2

Since the car starts from rest:
s = (1/2)at^2
s = (1/2) * 0.520 m/s^2 * (31.00 s)^2

Simplifying:
s = (1/2) * 0.520 m/s^2 * 961.00 s^2
s ≈ 249.32 m

So, the distance traveled by the race car is approximately 249.32 m (b) and the elapsed time is approximately 31.00 s (c).