91Ó°ÊÓ

Velocity of the car, $v=80\text{ k³¾}/\text{hr}$

Diameter of the tire, $r=66\text{ c³¾}$

The rate of change of displacement with respect to time gives the velocity and the rate of change of velocity with respect to time is called acceleration.

Before we go for calculation part, we will convert the$v\text{ a²Ô»å}\text{ }r$ into SI units.

$\begin{array}{c}v=80\text{ }\frac{\text{km}}{\text{hr}}\\ =80\text{ }\frac{\text{km}}{\text{hr}}×\frac{1000\text{″¾}}{1\text{ k³¾}}×\frac{1\text{ h°ù}}{3600\text{ s}}\\ =22.2\text{″¾}/\text{s}\end{array}$

$\begin{array}{c}r=66\text{cm}\\ =66\text{ }\text{cm}×\frac{0.01\text{″¾}}{1\text{ }\text{cm}}\\ =0.66\text{″¾}\end{array}$

In the cars reference frame, tire is performing only rotational motion. So center of the tire is at rest relative to the woman. So,$v_{\text{center}}=0\text{ }\text{m/s}$

Relative to woman, there is no translational motion of the tire, the motion is purely rotational.So the velocity of the top part of the tire is equal to the velocity which with any point of the tire is moving but it would be in the forward direction. The velocity of the point on the tire is determined in the above steps.

${\stackrel{→}{v}}_{top}=(22\text{″¾/²õ})\text{ }\widehat{i}$

The bottom point of the tire will have the same velocity as the top of the tire, only in the opposite direction

As the frame of reference is not accelerating, so the fixed points will have zero acceleration

In this the$\mathrm{Ó\neg }$(omega) is constant, which means the acceleration for points will be radial (centripetal),acting vertically downward.

,${\stackrel{→}{a}}_{top}=−\frac{v^2}{r}\widehat{j}$

$\begin{array}{c}{\stackrel{→}{a}}_{top}=−\frac{{(22\text{ ″¾}/\text{s})}^2}{0.33\text{″¾}}\widehat{j}\\ =(−1.5×{10}^3\text{ }{\text{m/s}}^{\text{2}})\widehat{j}\end{array}$

The acceleration will be same as that of at the top, acting vertically upward.

$\begin{array}{c}{\stackrel{→}{a}}_{bottom}=\frac{v^2}{r}\widehat{j}\\ =\frac{{(22\text{ ″¾}/\text{s})}^2}{0.33\text{″¾}}\widehat{j}\\ =(1.5×{10}^3\text{ }{\text{m/s}}^{\text{2}})\widehat{j}\end{array}$

Now, the car is having both translational and rotational motion,

${\text{v}}_{\text{center}}=22\text{ }\text{m/s}$

.${\stackrel{→}{v}}_{\text{center}}=(22\text{ }\text{m/s})\widehat{i}$

$\begin{array}{c}{v}_{top}=v_{rotational}+v_{translational}\\ =22\text{ ″¾}/\text{s}+22\text{ ″¾}/\text{s}\\ =+44\text{ }\text{m/s}\\ {\stackrel{→}{v}}_{\text{top}}=(+44\text{ }\text{m/s})\widehat{i}\end{array}$

The bottom of the tire is in firm contact with the road at any instant, so we have

$\begin{array}{c}{v}_{bottom}=v_{rotational}+v_{translational}\\ =−22\text{ ″¾}/\text{s}+22\text{ ″¾}/\text{s}\\ =0\text{ }\text{m/s}\end{array}$

As the frame of reference is not accelerating, so the fixed points will have zero acceleration

$a_{\text{center}}=0{\text{″¾/²õ}}^{\text{2}}$

As we are translating with the constant velocity, so translational acceleration will be zero, only rotational acceleration will be there,

$\begin{array}{c}{\stackrel{→}{a}}_{top}=−\frac{v^2}{r}\widehat{j}\\ =−\frac{{(22\text{ ″¾}/\text{s})}^2}{0.33\text{″¾}}\widehat{j}\\ =−(1.5×{10}^3\text{ }{\text{m/s}}^{\text{2}})\widehat{j}\end{array}$

The acceleration will be same as that of at the top,

$\begin{array}{c}{\stackrel{→}{a}}_{bottom}=\frac{v^2}{r}\widehat{j}\\ =\frac{{(22\text{ ″¾}/\text{s})}^2}{0.33\text{″¾}}\widehat{j}\\ =(1.5×{10}^3\text{ }{\text{m/s}}^{\text{2}})\widehat{j}\end{array}$

The acceleration at the bottom of the tire, relative to the hitchhiker is $(1.5×{10}^3\text{ }{\text{m/s}}^{\text{2}})\widehat{j}$.