Q18E - University Physics with Modern Physics

Question

The position of the front bumper of a test car under microprocessor control is given by $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{17}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{80}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{)}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{6}}\mathbf{)}{\mathbf{t}}^{\mathbf{6}}$. (a) Find its position and acceleration at the instants when the car has zero velocity. (b) Draw $\mathbf{x}\mathbf{-}\mathbf{t}\mathbf{,}\mathbf{\hspace{0.33em}}{\mathbf{v}}_{\mathbf{x}}\mathbf{-}\mathbf{t}\mathbf{\hspace{0.33em}}\mathbf{and}\mathbf{\hspace{0.33em}}{\mathbf{a}}_{\mathbf{x}}\mathbf{-}\mathbf{t}$ graphs for the motion of the bumper between $\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{and}\mathbf{\hspace{0.33em}}\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\hspace{0.33em}}\mathbf{s}$.

Step-by-Step Solution

Answer

a) the position and acceleration at the instants when the car has zero velocity are 2.17 m and $9.6m/s^2$ at t=0 and $38.4\hspace{0.33em}m/s^2$ and t=2 at respectively and b) the graphs has been provided below.

Step 1: Identification of the given data

The given data can be listed below as:

• The velocity of the car is zero.
• The time taken for the motion of the bumper are $t=0\hspace{0.33em}s\hspace{0.33em}and\hspace{0.33em}t=2\hspace{0.33em}s$ .

Step 2: Significance of the Newton’s first law for the test car

This law illustrates that a particular object will continue to be at rest or be at uniform motion unless an external force acts upon the particle.

Putting the value of the velocity in the equation of the position gives the position of the instant. Differentiating the given position of the test car gives the velocity and differentiating the gathered velocity and putting the required value provides the required acceleration.

Step 3: Determination of the position and acceleration of the car along with the graphs

a) The equation of the position of the position of the front bumper given in the problem can be stated as:

$x\left(t\right)=2.17\hspace{0.33em}m+(4.80\hspace{0.33em}m/s^2)t^2-(0.100\hspace{0.33em}m/s^6)t^6$… i)

Differentiating the above equation i), we get-

$v\left(t\right)=9.6t-6t^5$… ii)

whereis the velocity of the front bumper of the test car.

Differentiating equation ii). We get-

$a\left(t\right)=9.6-3t^4$

Where is the acceleration of the test car.

As the car has zero velocity at the instant, hence from the equation ii), we get-

$$\begin{gathered} 9.6t-6t^5=0 \\ t\left(9.6-6t^4\right)=0 \\ t=0\text{ and }2\mathrm{s} \end{gathered}$$

The two instants when the car has zero velocity are $t=0\hspace{0.33em}and\hspace{0.33em}2\hspace{0.33em}s$.

For t=0, the position of the car is-

$$\begin{gathered} x\left(0\right)=2.17+4.8\times (0{)}^2-0.100\times (0{)}^2 \\ x\left(0\right)=2.17\hspace{0.33em}m \end{gathered}$$

For t=0, the acceleration of the car is-

$$\begin{gathered} a\left(0\right)=9.6-3\times \left(0{)}^4 \\ a\right(0)=9.6\hspace{0.33em}m/s^2 \end{gathered}$$

For t=2, the position of the car is-

$$\begin{gathered} x2=2.17+4.8\times (2{)}^2-0.100\times (2{)}^2 \\ x\left(2\right)=15\hspace{0.33em}m \end{gathered}$$

For t=2s, the acceleration of the car is-

$$\begin{gathered} a\left(2\right)=9.6-3\times \left(2{)}^4 \\ a\right(0)=38.4\hspace{0.33em}m/s^2 \end{gathered}$$

Thus, the position and acceleration at the instants when the car has zero velocity are 2.17 m and $9.6m/s^2$ at t=0 s and 15 m and $38.4\hspace{0.33em}m/s^2$ at t=2 respectively.

b) the  x-t graph for the motion of the bumper between t=0s and t=2s is:

the $v_x-t$ graph for the motion of the bumper between is:

the $a_x-t$graph for the motion of the bumper between is: