A race car starts from rest on a circular track of radius 279 m. The car's speed increases at the constant rate of 0.760 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) elapsed time? s
To find the speed of the race car, we need to equate the magnitudes of the centripetal and tangential accelerations. The 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. The tangential acceleration is the rate at which the car's speed is increasing, given by:
at = 0.760 m/s^2
At the point where the magnitudes of the centripetal and tangential accelerations are equal, we can set ac equal to at:
v^2 / r = 0.760 m/s^2
Solving for v, we get:
v^2 = ac * r
v = √(0.760 m/s^2 * 279 m)
v ≈ 14.15 m/s
a) The speed of the race car is approximately 14.15 m/s.
To find the distance traveled, we can use the kinematic equation:
s = ut + (1/2)at^2
Since the car starts from rest, the initial velocity (u) is 0 m/s. The acceleration (a) is given as at, and we need to find the time (t) it takes for the magnitudes of the centripetal and tangential accelerations to become equal.
Using the equation above, we rearrange it to solve for t:
at^2 = 2s
t^2 = 2s / at
t = √(2s / at)
Plugging in the values:
t = √(2 * 279 m / 0.760 m/s^2)
t ≈ 21.1 s
b) The distance traveled by the car is approximately 279 m.
c) The elapsed time is approximately 21.1 s.
To solve this problem, we need to find the point where the magnitudes of the centripetal and tangential accelerations are equal. Let's assume that the car has reached a certain speed "v" at that point.
a) Speed of the race car:
Centripetal acceleration, ac = v^2 / r (where r is the radius of the circular track)
Tangential acceleration, at = v * t (where t is the time elapsed)
Since we want to find the point where both accelerations are equal, we can set them equal to each other:
v^2 / r = v * t
Solving for v:
v = r / t
Now, we need to determine the elapsed time t at this point. We know that the car's speed increases at a constant rate of 0.760 m/s^2. Therefore, the tangential acceleration is given as at = 0.760 m/s^2.
Substituting at = v * t in the equation above:
0.760 m/s^2 = (r / t) * t
Simplifying:
0.760 m/s^2 = r
Therefore, we can conclude that at the point where the magnitudes of the centripetal and tangential accelerations are equal, the speed of the race car is equal to the radius of the circular track, which is 279 m.
b) Distance traveled:
To find the distance traveled, we need to calculate the distance covered during the time it takes to accelerate to the speed of 279 m/s.
The final speed, vf = 279 m/s
The initial speed, vi = 0 m/s
Acceleration, a = 0.760 m/s^2
We can use the equation of motion:
vf^2 = vi^2 + 2a * d
Solving for d (distance traveled):
d = (vf^2 - vi^2) / (2a)
Substituting the given values:
d = (279^2 - 0^2) / (2 * 0.760)
Simplifying:
d = 384490.79 m
Therefore, the distance traveled by the race car is approximately 384491 m.
c) Elapsed time:
We can find the elapsed time by using the equation of motion:
vf = vi + a * t
Substituting the known values:
279 = 0 + 0.760 * t
Simplifying:
t = 366.447 s
Therefore, the elapsed time is approximately 366.447 seconds.