Q14E - University Physics with Modern Physics

Question

A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car’s motion, the eastward component of the car’s velocity is given by ${\mathbf{v}}_{\mathbf{x}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{860}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{3}}\mathbf{)}{\mathbf{t}}^{\mathbf{2}}$. What is the acceleration of the car when ${\mathbf{v}}_{\mathbf{x}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}$?

Step-by-Step Solution

Answer

The acceleration of the car, when $v_x=0.12\hspace{0.33em}m/s$ is $0.64\hspace{0.33em}m/s^2$.

Step 1: Identification of the given data

The given data is listed below as,

 

• The component in the east direction for the velocity of the car is, $v_x\left(t\right)=(0.860m/s^3)t^2$
• The velocity of the race car is, $v_x=0.12m/s$

Step 2: Use of velocity function for the determination of acceleration

It is known that the variation of the velocity with respect to time is known as acceleration. So, acceleration can be obtained by taking the derivative of the function of velocity with respect to time.

Step 3: Differentiation of function of velocity with respect to time

Take the derivative of the function of velocity with respect to time to find the value of acceleration.

$$\begin{gathered} a_x\left(t\right)=\frac{d}{dt}\left(v_x\left(t\right)\right) \\ =\frac{d}{dt}\left(\left(0.860\mathrm{m}/{\mathrm{s}}^3\right)t^2\right) \\ =\left(0.860\mathrm{m}/{\mathrm{s}}^3\right)2t \\ =\left(1.720\mathrm{m}/{\mathrm{s}}^3\right)\times t \end{gathered}$$           ……………………………………(1)

Step 4: Determination of time for the velocity

Substitute the value of velocity in the expression $v_x\left(t\right)=(0.860m/s^3)t^2$, and we get 

$$\begin{gathered} 0.12\mathrm{m}/\mathrm{s}=\left(0.860\mathrm{m}/{\mathrm{s}}^3\right){\mathrm{t}}^2 \\ \mathrm{t}=\sqrt{\frac{0.12\mathrm{m}/\mathrm{s}}{0.860\mathrm{m}/{\mathrm{s}}^2}} \\ =0.37\mathrm{s} \end{gathered}$$ 

Thus, the velocity will be $v_x=0.12\hspace{0.33em}m/s$  at $t=0.37s$ .

Step 5: Determination of acceleration at .

Substitute the value of t in equation (1), and we get.

$$\begin{gathered} a_x\left(3.735\mathrm{s}\right)=\left(1.72\mathrm{m}/{\mathrm{s}}^2\right)\times \left(0.37\mathrm{s}\right) \\ =0.64\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration at $v_x=0.12\hspace{0.33em}m/s$ is $0.64\hspace{0.33em}m/s^2$.