A race car starts from rest on a circular track of radius 507 m. The car's speed increases at the constant rate of 0.790 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the following.
(a) the speed of the race car
____ m/s
(b) the distance traveled
____ m
( c) the elapsed time
____ s
To solve this problem, we need to understand the concept of centripetal and tangential accelerations.
Centripetal acceleration (ac) is the acceleration that keeps an object moving in a circular path, and its magnitude can be found using the formula ac = v^2/r, where v is the velocity and r is the radius of the circular path.
Tangential acceleration (at) is the acceleration that changes the speed of an object moving in a circular path, and its magnitude can be found by taking the derivative of the speed with respect to time, at = dv/dt.
Now, let's solve the problem step by step.
(a) To find the speed of the race car when the magnitudes of centripetal and tangential accelerations are equal, we can equate the formulas for ac and at:
v^2/r = dv/dt
Since the car's speed is increasing at a constant rate, dv/dt is equal to the given constant rate of 0.790 m/s^2. Rearranging the equation, we get:
v^2 = r * dv/dt
v^2 = 507 * 0.790
Taking the square root of both sides gives us the speed of the race car:
v = sqrt(507 * 0.790)
v ≈ 20.002 m/s
So, the speed of the race car is approximately 20.002 m/s.
(b) To find the distance traveled, we need to calculate the time it takes for the magnitudes of centripetal and tangential accelerations to be equal. Since at = dv/dt, we can rewrite the equation as:
dv/dt = 0.790
Now, we can solve this differential equation to find the time. Integrating both sides, we get:
∫(1/v) dv = ∫0.790 dt
Taking the integral and applying the limits, we have:
ln(v) = 0.790t + C
Where C is the constant of integration. However, since the race car starts from rest (v = 0 at t = 0), we can determine that C = 0. Therefore, our equation becomes:
ln(v) = 0.790t
Exponentiating both sides, we get:
v = e^(0.790t)
Now, we can substitute the given value of v (20.002 m/s) to find the corresponding time t:
20.002 = e^(0.790t)
Taking the natural logarithm of both sides and isolating t, we can solve for t:
ln(20.002) = 0.790t
t = ln(20.002)/0.790
t ≈ 3.680 s
So, the elapsed time is approximately 3.680 s.
(c) The distance traveled can be found using the formula for distance (d) traveled in a circular path:
d = v*t
Substituting the values of v and t we found earlier:
d = 20.002 * 3.680
d ≈ 73.604 m
Therefore, the distance traveled is approximately 73.604 m.