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
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, we need to determine the point where the magnitudes of the centripetal and tangential accelerations are equal.
The centripetal acceleration is given by the formula ac = v^2 / r, where v is the speed and r is the radius of the circular track.
The tangential acceleration is given by the formula at = a, where a is the constant rate at which the car's speed increases, 8.80 m/s^2 in this case.
Setting these two accelerations equal, we have:
v^2 / r = a
Substituting the values given, we have:
v^2 / 4.30 = 8.80
Now, we can solve for the speed (v):
v^2 = 8.80 * 4.30
v^2 = 37.84
v ≈ √37.84
v ≈ 6.15 m/s
So, the speed of the race car is approximately 6.15 m/s.
To determine the elapsed time, we can use the kinematic equation:
v = u + at
Where u is the initial speed (0 m/s in this case), a is the acceleration (8.80 m/s^2), and t is the elapsed time.
Substituting the values, we have:
6.15 = 0 + 8.80t
Solving for t, we get:
t ≈ 0.70 seconds
So, the elapsed time is approximately 0.70 seconds.
To find the distance traveled, we can use the formula:
s = ut + 0.5at^2
Where s is the distance traveled, u is the initial speed (0 m/s), a is the acceleration (8.80 m/s^2), and t is the elapsed time (0.70 seconds).
Substituting the values, we have:
s = 0 + 0.5 * 8.80 * (0.70)^2
Simplifying, we get:
s ≈ 2.15 meters
So, the distance traveled is approximately 2.15 meters.