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 8.80 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, determine (a) the speed of the race car. b)the elapsed time c) the distance traveled

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To solve this problem, we need to consider the centripetal acceleration and the tangential acceleration of the race car.

Centripetal acceleration (ac) is given by the formula:
ac = v^2 / r
where v is the velocity of the car and r is the radius of the circular track.

Tangential acceleration (at) is given by the formula:
at = a
where a is the constant acceleration of the car.

(a) To find the speed of the race car, we need to equate the centripetal and tangential accelerations:
ac = at
v^2 / r = a

Rearranging the equation, we have:
v^2 = a * r

Substituting the given values:
v^2 = 8.80 m/s^2 * 4.30 m
v^2 = 37.84 m^2/s^2

Taking the square root of both sides:
v = √37.84 m/s
v ≈ 6.16 m/s

Therefore, the speed of the race car is approximately 6.16 m/s.

(b) To find the elapsed time, we can use the kinematic equation:
v = u + at
where u is the initial velocity, which is 0 m/s.

Rearranging the equation, we have:
t = (v - u) / a

Substituting the given values:
t = (6.16 m/s - 0 m/s) / 8.80 m/s^2
t ≈ 0.70 s

Therefore, the elapsed time is approximately 0.70 seconds.

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

Since the car starts from rest, its initial velocity (u) is 0 m/s.

Substituting the given values:
s = 0 m/s * 0.70 s + (1/2) * 8.80 m/s^2 * (0.70 s)^2
s ≈ 1.73 meters

Therefore, the distance traveled by the race car is approximately 1.73 meters.