Question

A car starts from rest and moves around a circular track of radius30.0M. Its speed increases at the constant rate of 0.500M/S². (a) What is the magnitude of its net linear acceleration 15.0slater? (b) What angle does this net acceleration vector make with the car’s velocity at this time?

Short answer

1. The magnitude of net linear acceleration 15s later is, $\left|\stackrel{\rightharpoonup }{a}\right|=1.94m/s^2$
2. The angle this net acceleration vector makes with the car’s velocity is, $\theta =75.1^{\circ }$.

Step-by-step solution

Understanding the given information

1. The radius is, r = 30.0m.
2. The constant acceleration of car is, a = 0.500m/s².
3. Time is t = 15.0s.

Concept and Formula used for the question

By using kinematic equation, we can find the linear velocity v.By usingthis v, we can find the angular velocity $\mathit{\omega }$and angular acceleration $\mathit{\alpha }$, and using all these values, we can find the magnitude of net linear acceleration15 slater and theangle this net acceleration vector makes with the car’s velocity.

1. The kinematic equation is $\mathit{v}\mathbf{=}{\mathit{v}}_{\mathbf{0}}\mathbf{+}\mathit{a}\mathit{t}$
2. The angular velocity is $\mathit{\omega }\mathbf{=}\frac{\mathbf{v}}{\mathbf{r}}$
3. The angular acceleration is $\mathit{\alpha }\mathbf{=}\frac{\mathbf{\omega }}{\mathbf{t}}$
4. The tangential acceleration is${\mathit{a}}_{\mathbf{t}}\mathbf{=}\mathit{\alpha }\mathit{r}$
5. The radial acceleration is ${\mathit{a}}_{\mathbf{r}}\mathbf{=}{\mathit{\omega }}^{\mathbf{2}}\mathit{r}$
6. The magnitude of acceleration is role="math" localid="1661063908108" $\mathbf{\left|}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}\mathbf{\right|}\mathbf{=}\sqrt{({a_r}^2+{a_t}^2)}$
7. The angle this net acceleration vector makes with the car’s velocity is $\mathit{\theta }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{a_r}{a_t}\right)$

(a) Calculation for the magnitude of net linear acceleration  later

We know that the kinematic equation is

$v=v_0+at$

But $v_0=0$, so the linear velocity v is given by

$v=at$

Substitute the all the value in the above equation.

$\begin{array}{rcl}v& =& 0.500m/s^2\times 15s\\ & =& 7.50m/s\end{array}$

The angular velocity $\omega$is given by

$\omega =\frac{v}{r}$

Substitute the all the value in the above equation.

$\begin{array}{rcl}\omega & =& \frac{7.50m/s}{30.0m}\\ & =& 0.250rad/s\end{array}$

Now, the angular acceleration is given by

$\alpha =\frac{\omega }{t}$

Substitute the all the value in the above equation.

$\begin{array}{rcl}\alpha & =& \frac{0.250rad/s}{15s}\\ & =& 0.017rad/s^2\end{array}$

The radial acceleration is given by

$a_r={\omega }^{2}r$

Substitute the all the value in the above equation.

$\begin{array}{rcl}{a}_r& =& {\left(0.250rad/s^2\right)}^2\times 30.0m\\ & =& 1.875m/s^2\end{array}$

The tangential acceleration is given by

$a_t=\alpha r$

Substitute the all the value in the above equation.

$\begin{array}{rcl}{a}_t& =& 0.0166rad/s^2\times 30.0m\\ & =& 0.50m/s^2\end{array}$

Thus, the magnitude of net linear acceleration is

$\left|\stackrel{\rightharpoonup }{\mathrm{a}}\right|=\sqrt{({{\mathrm{a}}_{\mathrm{r}}}^2+{{\mathrm{a}}_{\mathrm{t}}}^2)}$

Substitute the all the value in the above equation.

$\begin{array}{rcl}\left|\stackrel{\rightharpoonup }{\mathrm{a}}\right|& =& \sqrt{{\left(1.875\mathrm{m}/{\mathrm{s}}^2\right)}^2+{\left(0.50\mathrm{m}/{\mathrm{s}}^2\right)}^2}\\ & =& 1.94m/s^2\end{array}$

Hence the linear acceleration is, 1.94m/s².

(b) Calculation for the angle the net acceleration vector makes with the car’s velocity at this time

The angle this net acceleration vector makes with the car’s velocity at this time is

$\theta ={\tan }^{-1}\left(\frac{a_r}{a_t}\right)$

Substitute the all the value in the above equation.

$\begin{array}{rcl}\theta & =& {\tan }^{-1}\left(\frac{1.875m/s^2}{0.500m/s^2}\right)\\ & =& 75.1^o\end{array}$

Hence the angle is, 75.1^o.