A car that is initially at rest moves along a circular path with a constant tangential acceleration component of 2.07 m/s2. The circular path has a radius of 42.4 m. The initial position of the car is at the far west location on the circle and the initial velocity is to the north.
(a) After the car has traveled 1/4 of the circumference, what is the speed of the car?
1 m/s
(b) At this point, what is the radial acceleration component of the car?
2 m/s2
(c) At this same point, what is the total acceleration of the car?
magnitude 3 m/s
direction 4 ° east of south
To solve this problem, we'll need to break it down into several steps.
(a) To find the speed of the car after it has traveled 1/4 of the circumference, we can use the equations of circular motion. First, we need to find the distance traveled by the car when it reaches the point after 1/4 of the circumference.
The formula to calculate the circumference of a circle is C = 2πr, where r is the radius. In this case, the radius is given as 42.4 m. So, the circumference of the circle is C = 2π(42.4) = 266.56 m.
1/4 of the circumference would be (1/4)(266.56) = 66.64 m.
Now, we can use the formula for distance traveled in circular motion, s = rθ, where s is the distance, r is the radius, and θ is the angle in radians.
Since the car has traveled 66.64 m, we can rearrange the formula to solve for θ:
θ = s / r
θ = 66.64 m / 42.4 m
θ ≈ 1.573 radians
Now, we can use the formula for speed in circular motion, v = rω, where v is the speed, r is the radius, and ω is the angular velocity.
Since we have the angle in radians, we can use the equation ω = ω0 + αt, where ω0 is the initial angular velocity, α is the tangential acceleration, and t is the time.
Since the car starts from rest (initial velocity is zero), ω0 = 0. Also, the tangential acceleration is given as 2.07 m/s^2.
ω = αt
ω = (2.07 m/s^2)(t)
To find t, we can use the equation s = vt, where s is the distance traveled (66.64 m) and v is the speed we want to find. Rearranging the equation, we get t = s / v.
Plugging in the values, we have:
1.573 radians = (2.07 m/s^2)(s / v)
v ≈ 1.003 m/s
So, the speed of the car after it has traveled 1/4 of the circumference is approximately 1 m/s.
(b) The radial acceleration component represents the acceleration towards the center of the circle. In this case, the radial acceleration component is equal to the centripetal acceleration.
The formula for centripetal acceleration is ac = v^2 / r, where v is the speed and r is the radius.
Plugging in the values, we have:
ac = (1.003 m/s)^2 / 42.4 m
ac ≈ 0.0235 m/s^2
So, the radial acceleration component of the car at this point is approximately 0.0235 m/s^2.
(c) The total acceleration of the car can be calculated by combining the tangential and radial acceleration components.
We can use the Pythagorean theorem to find the magnitude of the total acceleration (a) using the equation:
a^2 = at^2 + ar^2
Plugging in the values, we have:
a^2 = (2.07 m/s^2)^2 + (0.0235 m/s^2)^2
Simplifying, we get:
a ≈ 2.071 m/s^2
So, the magnitude of the total acceleration of the car at this point is approximately 2.071 m/s^2.
To find the direction, we can use the trigonometric relationship between the tangential and radial acceleration components. The angle (θ) between the resultant acceleration vector and the north direction can be found using the equation:
tan(θ) = ar / at
Plugging in the values, we have:
tan(θ) = 0.0235 m/s^2 / 2.07 m/s^2
θ ≈ 0.0114 radians
Converting radians to degrees, we have:
θ ≈ 0.0114 radians × (180° / π radians)
θ ≈ 0.6537°
Since the angle is east of south, we subtract this angle from the south direction:
θ = 180° - 0.6537°
θ ≈ 179.3463°
So, the direction of the total acceleration of the car at this point is approximately 179.3463° east of south.