# The tires of a car make 64 revolutions as the car reduces its speed uniformly from 95.0 km/h to...

## Question:

The tires of a car make 64 revolutions as the car reduces its speed uniformly from 95.0 km/h to 62.0 km/h. The tires have a diameter of 0.80.

If the car continues to decelerate at this rate, how much more time is required for it to stop?

## Rotational motion:

Just like equations of motion in linear motion, we have equations of motion in rotational motion as well.

These equations describe the angular position, angular velocity and angular acceleration of any object rotating about an axis of rotation.

The equations of motion in rotational motion are;

• {eq}\omega = \omega_0 + \alpha t {/eq}
• {eq}\theta =\theta_0 + \omega_0 t +\dfrac{1}{2}\alpha t^2 {/eq}
• {eq}\omega^2 = \omega_0^2 + 2 \alpha (\theta-\theta_0) {/eq}

where,

• {eq}\omega_0 {/eq} is the initial angular velocity
• {eq}\omega {/eq} is the angular velocity at any time t.
• {eq}\alpha {/eq} is the constant angular acceleration
• {eq}\theta_0 {/eq} is the initial angular displacement
• {eq}\theta {/eq} is the angular displacement at any time t.

## Answer and Explanation:

Given:

• The initial speed of a car is {eq}v_1 = 95.0 \ km/h = 95.0 \dfrac {1000} {3600} \ m/s =26.4 \ m/s {/eq}.
• The final speed of a car is {eq}v_2 = 62.0 \ km/h = 62.0 \dfrac {1000} {3600} \ m/s = 17.2 \ m/s {/eq}.
• The number of revolutions of tires of a car is {eq}N = 64 \ rev. {/eq}.
• The radius of a tire of a car is {eq}r = \dfrac {d}{2} = \dfrac {0.80}{2} = 0.40 \ m {/eq}.

Let

• The angular deceleration of a car is {eq}\alpha {/eq}
• The time is required to stop a car is {eq}t {/eq}
• The angular displacement of the car is {eq}\Delta \theta {/eq}
• The initial angular speed of the car is {eq}\omega_1 {/eq}
• The final angular speed of the car is {eq}\omega_2 {/eq}

The angular displacement of the car is

{eq}\begin{align} \Delta \theta & = 2\pi N \\ & = 2\pi \times 64 \\ \implies \Delta \theta & = 402.3 \ rad \\ \end{align} {/eq}

The initial angular speed of a car is

{eq}\begin{align} \omega_1 &= \dfrac {v_1}{r} \\ &=\dfrac {26.4 }{0.40} \\ \implies \omega_1 &= 65.97 \ rad/s \\ \end{align} {/eq}.

The final angular speed of a car is

{eq}\begin{align} \omega_2 &= \dfrac {v_2}{r} \\ &=\dfrac {17.2 }{0.40} \\ \implies \omega_2 &= 43.1 \ rad/s \\ \end{align} {/eq}.

The equation of motion in rotational motion is;

{eq}\begin{align} \omega_2^2 = \omega_1^2 + 2 \alpha (\Delta \theta) \\ (43.1)^2 & = (65.97)^2 + 2 \alpha (402.3) \\ 1,857.61 & = 4,352.04 + 804.6\alpha \\ -2,494.43 & = 804.6\alpha \\ \implies \alpha &=- 3.1 \ rad/s^2 \\ \end{align} {/eq}

If the car continues to decelerate at {eq}\alpha =- 3.1 \ rad/s^2 {/eq} so, the time required for car to stop is given by:

{eq}\begin{align} \omega & = \omega_0 + \alpha t \\ \omega_f & = \omega_2 + \alpha t \\ 0 & = 43.1 -3.1t \\ 3.1t & = 43.1 \\ \implies t & = 13.9 \ s \\ \end{align} {/eq}

Therefore, the time required for the car to stop is {eq}13.9 \ s {/eq}.

#### Learn more about this topic:

Practice Applying Rotational Motion Formulas

from Physics 101: Help and Review

Chapter 17 / Lesson 15
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