Q17E - University Physics with Modern Physics

Question

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 for $\mathbf{t}=\mathbf{0} \mathbf{to} \mathbf{t}=\mathbf{5}.\mathbf{00} \mathbf{s}$. 

 (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$\mathbf{t}=\mathbf{0} \mathbf{to} \mathbf{t}=\mathbf{5}.\mathbf{00} \mathbf{s}$.

Step-by-Step Solution

Answer

a) the average acceleration of the car is ${\text{0.5m/s}}^2$ and

 b) the instantaneous acceleration of the car for$t=\text{0} \text{and} t=\text{5.00} \text{s}$ is $0m/s^2 and 5m/s^2$respectively and

 c) the$v_x-t \text{and} a_x-t$ graph has been illustrated below.

Step 1: Identification of the given data

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$\text{t}=\text{0} \text{and} \text{t}=\text{5.00} \text{s}$ .

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

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.

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

$=0.5m/s^2$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:

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

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

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

Thus, the instantaneous acceleration of the car at t=0 and t=5.00s is$0m/s^2 and 5m/s^2$ , 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: