91Ó°ÊÓ

The given data can be listed below as:

• At the beginning, the time is, 0s
• At the beginning, the acceleration is,$5.95{\text{m/s}}^{\text{2}}$
• At the first stage, the time is, 1.00s
• At the first stage, the acceleration is,$5.52{\text{m/s}}^{\text{2}}$
• At the second stage, the time is,2.00s
• At the second stage, the acceleration is,$5.08{\text{m/s}}^{\text{2}}$
• At the third stage, the time is,3.00s
• At the third stage, the acceleration is,$4.55{\text{m/s}}^{\text{2}}$
• At the fourth stage, the time is,4.00s
• At the fourth stage, the acceleration is,${\text{3.96m/s}}^{\text{2}}$
• At the fifth stage, the time is,5.00s

At the fifth stage, the acceleration is,${\text{3.40m/s}}^{\text{2}}$

The velocity of an object is described as the rate of change of displacement with time. The differentiation of velocity with respect to time is described as the acceleration of an object.

The graph of the acceleration versus time has been described below:

In the above graph, it has been observed that with the increase in the time, the acceleration is decreasing continuously in a straight line. With the help of the graph, the equation of the straight line can be identified.

The equation of the straight line can be expressed as:

$a\left(t\right)=\left(-0.5131{\text{m/s}}^{\text{3}}\right)t+\left(6.0262{\text{m/s}}^{\text{2}}\right)$ …(¾±)

Here, is the acceleration with respect to time and is the time taken by the model car.

Thus, the equation of the straight line that gives the best fit to the data is $a\left(t\right)=\left(-0.5131{\text{m/s}}^{\text{3}}\right)t+\left(6.0262{\text{m/s}}^{\text{2}}\right)$.

The equation (i) has been recalled below:

$a\left(t\right)=\left(-0.5131{\text{m/s}}^{\text{3}}\right)t+\left(6.0262{\text{m/s}}^{\text{2}}\right)$

Integrating the above equation with respect to the time to get the value of the velocity.

$$\begin{gathered} v\left(t\right)=∫a\left(t\right)dt \\ =∫\left[\left(-0.5131{\text{m/s}}^{\text{3}}\right)t+\left(6.0262{\text{m/s}}^{\text{2}}\right)\right]dt \\ =-0.5131\frac{t^2}{2}+6.0262t+C \\ =-0.25655t^2+6.0262t+C \end{gathered}$$

…(¾±¾±)

Here, C is a constant.

At the initial condition, the velocity at the time t =0 is 0 .

Substitute 0 for t in the above equation.

$$\begin{gathered} v\left(0\right)=-0.25655×{\left(0\right)}^2+6.0262×\left(0\right)+C \\ 0=-0.25655×{\left(0\right)}^2+6.0262×\left(0\right)+C \\ C=0 \end{gathered}$$

Substitute 0 for C in the above equation.

$v\left(t\right)=-0.25655t^2+6.0262t$ …(¾±¾±¾±)

The graph of v versus t has been provided below:

Here, in the above graph, it has been identified that with the increase in the time, the velocity also increases but not in a constant pace. Hence, the graph is not a straight line.

Thus, the speed of the car as a function of time is $v\left(t\right)=-0.25655t^2+6.0262t$.

The graph is not a straight line.