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 determine the speed of the race car at the point where the magnitudes of the centripetal and tangential accelerations are equal, we can follow these steps:

1. Identify the given values:
- Radius of the circular track, r = 4.30 m
- Constant rate of the car's speed increasing, a = 2.90 m/s²

2. Calculate the centripetal acceleration:
The centripetal acceleration (ac) can be calculated using the formula:
ac = v² / r, where v is the speed of the car.
Since the car starts from rest, the initial speed (v₀) is 0, and thus, the centripetal acceleration simplifies to:
ac = v² / r = (v - v₀)² / r = v² / r

3. Calculate the tangential acceleration:
The tangential acceleration (at) can be calculated using the formula:
at = a
The tangential acceleration is given as the constant rate at which the car's speed increases.

4. Set the centripetal and tangential accelerations as equal:
ac = at
v² / r = a
Solve this equation for v (car's speed).

5. Rearrange the equation to solve for v:
v² = a * r
v = √(a * r)

6. Substitute the given values and calculate:
v = √(2.90 m/s² * 4.30 m)
v ≈ 3.82 m/s

Therefore, at the point where the magnitudes of the centripetal and tangential accelerations are equal, the speed of the race car is approximately 3.82 m/s.