A race car, starting from rest, travels around a circular turn of radius 20.8 m. At a certain instant, the car is still accelerating, and its angular speed is 0.515 rad/s. At this time, the total acceleration (centripetal plus tangential) makes an angle of 35.1° with respect to the radius. What is the magnitude of the total acceleration?
The ratio of centripetal to tangential acceleration is the tangent of 35.1 degrees. You can compute the centripetal accleration (V^2/R) and get the tangential and magnitude from the angle.
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To find the magnitude of the total acceleration, we can split it into two components: the centripetal acceleration and the tangential acceleration.
The centripetal acceleration represents the acceleration towards the center of the circular path. It is given by the formula:
ac = v^2 / r
where v is the linear speed and r is the radius of the circle.
In this case, we are given the angular speed (ω) of the car, but we can convert it to linear speed (v) using the formula:
v = ω * r
So, substituting the values given, we have:
v = 0.515 rad/s * 20.8 m = 10.712 m/s
Now, we can calculate the centripetal acceleration:
ac = (10.712 m/s)^2 / 20.8 m = 5.49 m/s^2
The tangential acceleration represents the change in linear speed of the object as it moves along the circular path. It can be calculated using the formula:
at = α * r
where α is the angular acceleration. However, in this case, we are not given the angular acceleration directly. Instead, we are given the angle between the total acceleration and the radius.
To find the tangential acceleration, we need to use trigonometry. Since the total acceleration and the tangential acceleration form a right triangle, we can use the sine function:
sin(θ) = at / a
where θ is the angle between the total acceleration and the radius.
Rearranging the formula, we get:
at = a * sin(θ)
Substituting the known values, we have:
at = 5.49 m/s^2 * sin(35.1°) = 3.14 m/s^2
Now, we can find the magnitude of the total acceleration. The total acceleration is the hypotenuse of the right triangle formed by the centripetal acceleration and the tangential acceleration. We can use the Pythagorean theorem to find it:
a = sqrt(ac^2 + at^2)
Substituting the known values, we have:
a = sqrt((5.49 m/s^2)^2 + (3.14 m/s^2)^2) ≈ 6.48 m/s^2
Therefore, the magnitude of the total acceleration is approximately 6.48 m/s^2.