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A car accelerates uniformly from rest and covers a distance of 79 m in 8.6 s. ...
Mar 19, 2024
A car accelerates uniformly from rest and covers a distance of 79 m in 8.6 s. If the diameter of a tire is 44 cm Find the final angular velocity of one of the car’s wheels
Solution
a
Calculate the linear acceleration of the car: a=vf−vita = \frac{v_f - v_i}{t} where viv_i is the initial velocity, vfv_f is the final velocity, and tt is the time. Since the car starts from rest, vi=0v_i = 0. We need to find vfv_f and we have t=8.6t = 8.6 s
b
Use the equation of motion for constant acceleration: s=vit+12at2s = v_i t + \frac{1}{2} a t^2 where ss is the distance covered. We know s=79s = 79 m and vi=0v_i = 0, so we can solve for aa
c
Calculate the linear acceleration: a=2st2=2×79 m(8.6 s)2a = \frac{2s}{t^2} = \frac{2 \times 79 \text{ m}}{(8.6 \text{ s})^2}
d
Convert the diameter of the tire to radius in meters: r=d2=44 cm2=22 cm=0.22 mr = \frac{d}{2} = \frac{44 \text{ cm}}{2} = 22 \text{ cm} = 0.22 \text{ m}
e
Use the relationship between linear velocity and angular velocity: v=rωv = r\omega to find the final angular velocity ω\omega
f
Calculate the final linear velocity of the car using the acceleration: vf=vi+at=0+a×8.6 sv_f = v_i + at = 0 + a \times 8.6 \text{ s}
g
Substitute the calculated acceleration into the equation for vfv_f: vf=a×8.6 sv_f = a \times 8.6 \text{ s}
h
Calculate the final angular velocity: ω=vfr\omega = \frac{v_f}{r}
Answer
The final angular velocity of one of the car’s wheels is ω\omega.
Key Concept
The relationship between linear acceleration and angular velocity in circular motion is crucial. The linear acceleration aa can be used to find the final linear velocity vfv_f, which is then related to the angular velocity ω\omega through the equation v=rωv = r\omega, where rr is the radius of the wheel.
Explanation
By using the equations of motion to find the car's linear acceleration and velocity, and knowing the radius of the wheel, we can calculate the final angular velocity of the wheel. This demonstrates the connection between linear and angular quantities in rotational motion.
instead of using the kinematics equations, could you tell me if using a = (vf-v0)/t and v=(xf-x0)/t is fine
Solution
a
Calculate average acceleration using a=vf−v0ta = \frac{v_f - v_0}{t}
b
Calculate final velocity using v=xf−x0tv = \frac{x_f - x_0}{t}
c
Since the car starts from rest, v0=0v_0 = 0, and the final velocity vfv_f can be found from step b
d
Substitute vfv_f into the equation from step a to find the average acceleration aa
e
Use the average acceleration to find the final angular velocity of the wheel using the relationship ω=vfr\omega = \frac{v_f}{r}, where rr is the radius of the tire
Answer
Using the equations a=vf−v0ta = \frac{v_f - v_0}{t} and v=xf−x0tv = \frac{x_f - x_0}{t} is fine for calculating the final angular velocity of the car's wheel, as long as the acceleration is uniform.
Key Concept
Kinematic equations and the relationship between linear and angular quantities in uniform circular motion.
Explanation
The equations a=vf−v0ta = \frac{v_f - v_0}{t} and v=xf−x0tv = \frac{x_f - x_0}{t} are valid for uniformly accelerated motion, which allows us to find the final velocity and acceleration. These can then be used to find the angular velocity of the wheel.
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