A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a constant acceleration of$20m/s^2$, and the car has an acceleration of$\mathbf{3}\mathbf{.}\mathbf{40}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. The car overtakes the truck after the truck has moved$60.0m$. (a) How much time does it take the car to overtake the truck? (b) How far was the car behind the truck initially? (c) What is the speed of each when they are abreast? (d) On a single graph, sketch the position of each vehicle as a function of time. Take$x=0$at the initial location of the truck.

The given data can be listed below as,

• The acceleration of the truck is$2.10{\text{m/s}}^{\text{2}}$ .
• The acceleration of the car is$3.40{\text{m/s}}^{\text{2}}$ .
• The truck has moved about$60.0\text{m}$ .

This law states that a particular object will continue to move in a uniform motion unless the object is resisted by an external force.

The equation of displacement for the car and the truck gives the time and the distance of the car. Moreover, the equation of velocity gives the speed of the truck and the car.

The free body diagram of the car and the truck has been illustrated below-

From the above figure, it has been identified that are the acceleration of the car and the truck respectively.

a)

From Newton’s first law, the equation of motion of the truck has been provided below-

$S_T=\frac{a_T{t}^2}{2}$

Here,$s_T$is the distance moved by the truck which is$60.0\text{m}$and$a_T$is the acceleration of the truck that is $2.10{\text{m/s}}^{\text{2}}$and t is the time taken by the car.

Substituting the values in the above equation, we get-

$$\begin{gathered} \text{60.0}\text{m =}\frac{\left(\text{2.10}{\text{m/s}}^{\text{2}}\right).t^2}{\text{2}} \\ {t}^2=57.14{\text{s}}^{\text{2}} \\ t=7.559\text{s} \end{gathered}$$

Thus, the car takes about $7.559\text{s}$to overtake the truck

b)

From Newton’s first law, the equation of motion of the car has been provided below-

$S_C=\frac{a_C{t}^2}{2}$

Here, $s_c$is the distance moved by the car, and$a_c$is the acceleration of the car that is $3.40{\text{m/s}}^{\text{2}}$andt is the time taken by the car that is$7.559\text{s}$.

Substituting the values in the above equation, we get-

$$\begin{gathered} {\text{S}}_C\text{=}\frac{\left(\text{3.40}{\text{m/s}}^{\text{2}}\right).{\left(7.559\text{s}\right)}^2}{\text{2}} \\ {s}_C=97.135\text{m} \end{gathered}$$

Hence, the car was behind the truck at a distance of:

$\begin{array}{rcl}d& =& s_C-s_T\\ & =& 97.135\text{m - 60}\text{m}\\ & & \text{= 37.135m}\\ & & \end{array}$

Thus, the car was behind the truck at a distance of $\text{37.135m}$.

c)

From Newton’s first law, the velocity of the truck is expressed as:

$V_T=a_{T}t$(As the initial velocity of the truck is zero)

Substituting the values in the above equation, we get-

$$\begin{gathered} v_T=\left(2.10{\text{m/s}}^{\text{2}}\right)\times \left(7.559\text{s}\right) \\ {v}_T=15.8739\text{m/s} \end{gathered}$$

From Newton’s first law, the velocity of the car is expressed as:

$V_C=a_{C}t$(As the initial velocity of the truck is zero)

Substituting the values in the above equation, we get-

$$\begin{gathered} v_C=\left(3.40{\text{m/s}}^{\text{2}}\right)\times \left(7.559\text{s}\right) \\ {v}_C=25.7006\text{m/s} \end{gathered}$$

Thus, the speed of the truck and the car arelocalid="1655193921611" $\text{15.8739}\text{m/s}\text{and}\text{25.7006}\text{m/s}$respectively.

d)

The graph of the position of each vehicle has been illustrated below-