A toy race car starts from rest on a circular track of radius 4.30 m. The cars speed increases at the constant rate of 2.90 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, determine (a) the speed of the race car.

R=4.3 m, a(τ)=8.8 m/s².

a(τ)=a(c) =v²/R =>
v=sqrt{a(c) •R}= sqrt{a(τ) •R}=
=sqrt{8.8•4.3}=6.15 m/s.

v=a(τ) •t,
t= v/a(τ)=6.15/8.8 =0.699 s= 0.7 s.

s= a(τ) •t²/2 =8.8•0.7²/2=2.156 m.

To find the speed of the race car at the point where the magnitudes of the centripetal and tangential accelerations are equal, we need to determine the magnitudes of these accelerations first.

1. Centripetal acceleration (ac):
Centripetal acceleration is given by the formula ac = v^2 / r, where v is the speed of the car and r is the radius of the circular track. Rearranging this formula, we have v = sqrt(ac * r).

Given that the radius of the circular track is 4.30 m, we can substitute this value into the equation.

v = sqrt(ac * 4.30)

2. Tangential acceleration (at):
Tangential acceleration can be calculated using the formula at = a, where a is the constant rate at which the car's speed increases.

Given that the car's speed increases at the constant rate of 2.90 m/s^2, we can substitute this value into the equation.

at = 2.90

3. Setting the magnitudes of centripetal and tangential acceleration equal:
Since we know the value of tangential acceleration (at = 2.90), we need to find the value of centripetal acceleration (ac) at the same point.

Since the magnitudes of centripetal and tangential accelerations are equal at this point, we can equate their values:

ac = at

4. Solving for v:
Now we can substitute the given value of tangential acceleration (at = 2.90) into the equation for centripetal acceleration:

sqrt(ac * 4.30) = 2.90

Squaring both sides of the equation, we get:

ac * 4.30 = 2.90^2

ac * 4.30 = 8.41

Dividing both sides of the equation by 4.30, we find:

ac = 8.41 / 4.30

ac ≈ 1.95 m/s^2

Now, we can substitute this value for ac back into the equation for speed (v = sqrt(ac * r)), along with the given value for the radius (r = 4.30 m):

v = sqrt(1.95 * 4.30)

v ≈ sqrt(8.39)

v ≈ 2.90 m/s

Therefore, the speed of the car at the point where the magnitudes of the centripetal and tangential accelerations are equal is approximately 2.90 m/s.