Question: A car’s velocity as a function of time is given by${\mathbf{v}}_{\mathbf{x}}\left(\mathbf{t}\right)=\alpha +\beta {\mathbf{t}}^2$, where$\alpha =\mathbf{3}.\mathbf{00}m/s$and $\beta =\mathbf{0}.\mathbf{100}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{3}}$.(a) Calculate the average acceleration for the time interval $\mathbf{t}=\mathbf{0}\mathbf{to}\mathbf{t}=\mathbf{5}.\mathbf{00}\mathbf{s}$. (b) Calculate the instantaneous acceleration forrole="math" $\mathbf{t}=\mathbf{0}\mathbf{to}\mathbf{t}=\mathbf{5}.\mathbf{00}\mathbf{s}$.

(c) Draw ${\mathbf{v}}_{\mathbf{x}}\text{-}\mathbf{t}\mathbf{and}{\mathbf{a}}_{\mathbf{x}}\text{-}\mathbf{t}$graphs for the car’s motion between$\mathbf{t}=\mathbf{0}\mathbf{to}\mathbf{t}=\mathbf{5}.\mathbf{00}\mathbf{s}$.

The given data can be listed below as:

• The value of is given as$3.00m/s$.
• The value of is given as$0.100m/s^3$$0.100m/s^3$.
• The time interval ranges from$\text{t}=\text{0}\text{and}\text{t}=\text{5.00}\text{s}$.

This law states that an object will continue to move in uniform motion unless an external force acts upon the object.

Differentiating the equation of motion of the car along a straight line and putting the value of the time will give the average and the instantaneous acceleration of the car which will help to find out the graph.

$=0.5m/s^2$According to the question, the car’s velocity as a function of time is expressed as:

$v_x\left(t\right)=\alpha +\beta t^2$

Here, $v_x\left(t\right)$is the velocity of the car as a function of time

a) The formula of the average acceleration for the car is expressed as:

$a_{x,avg}=\frac{v_{x2}-v_{x1}}{t_2-t_1}$

Here, is the velocity at$t=5.00s$which is$\text{3.00m/s+}\left(0.100m/s^3\right)\text{×}{\left(5s\right)}^2\text{=5.5m/s}$, $v_{x1}$is the velocity at $t=0$is ,$\text{3.00m/s+}\left({\text{0.100m/s}}^3\right)\text{×}{\left(\text{0}\text{s}\right)}^2\text{\$=3.0m/s}$and the value of $t_2$and $t_1$are 5s and 0 s respectively.

Substituting the values in the above equation, we get-

$\begin{array}{rcl}{a}_{x,avg}& =& \frac{\text{5.5m/s-3.0m/s}}{5s-0s}\\ & =& 0.5m/s^2\end{array}$

Thus, the average acceleration of the car is $0.5m/s^2$.

b) The formula of instantaneous acceleration for the car is expressed as:

$a_{inst}=\frac{d}{dt}\left(v\left(t\right)\right)$

Here, $a_{inst}$is the instantaneous acceleration of the car and$\frac{d}{dt}\left(v\left(t\right)\right)$is the differentiation of the velocity function with respect to time.

The equation of the instantaneous acceleration is:

${\alpha }_{inst}=2\beta t$

The instantaneous acceleration at$t=0$is given by:

$\begin{array}{rcl}{a}_{instatt=0s}& =& 2\times \left(\text{0.100m/}{s}^3\right)\times {\left(0s\right)}^2\\ & =& 0m/s^2\end{array}$

The instantaneous acceleration at$t=5s$is given by:

$$\begin{gathered} a_{instatt=5s}=2\times \left(\text{0.100m}{\text{/s}}^3\right)\times {\left(5s\right)}^2 \\ =5m/s^2 \end{gathered}$$

Thus, the instantaneous acceleration of the car at t=0 and t=5.00s is$0m/s^{2}and5m/s^2$, respectively.

c) The$v_x-t$graph for the motion of the car for t=0 and t=5.00 is expressed as:

The $a_x-t$graph for the motion of the car for$t=0\text{s}\text{and}t=5.00\text{s}$is expressed as: