91影视

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,$v_x=0.12m/s$

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.

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

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

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$v_x=0.12鈥�m/s$ at$t=0.37s$ .

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

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

Thus, the acceleration at$v_x=0.12鈥�m/s$ is$0.64鈥�m/s^2$.