91Ó°ÊÓ

The given data can be listed below as,

• The speed of the truck is $20.0m/s$.
• The car is also travelling at $20.0m/s$.
• The truck’s front bumper is$24.0m$ behind.
• The car accelerates at a rate of $0.600m/s^2$.
• The rear of the car is about$26.0m$ .
• The length of the car is $4.5m$.
• The length of the truck is$21.0m$ long.

This law states the motion of an object can be changed with the influence of an external force, otherwise it will continue to move like it was moving previously.

The equation of displacement gives the time required and the distance for the car and the equation of velocity gives the final velocity of the car.

a)

Due to the front bumper, the length of the truck and the car and the front of the truck, the car needs to move to a distance of:

$\begin{array}{rcl}l& =& l_1+l_2+l_3+l_4\\ & =& 24\text{m + 26}\text{m}+21\text{m}+4.5\text{m}\\ & =& 75.5\text{m}\\ & & \end{array}$

As both the vehicles has the same speed, hence, from the Newton’s first law, the time taken by the car is expressed as:

$s=ut+1/2a{t}^2$

Here, s is the displacement of the car that is$75.5m$ , u is the initial velocity of the car that is 0 and a is the acceleration of the car that is$0.6m/s^2$ .

Substituting the values in the above equation, we get-

$$\begin{gathered} t=\left(\sqrt{\frac{2.s}{a}}\right) \\ t=\left(\sqrt{\frac{\text{2}×\text{75.5}\text{m}}{\text{0.6}{\text{m/s}}^{\text{2}}}}\right) \\ t≈15.86 \text{s} \end{gathered}$$

Thus, time required by the car to pass the truck is .

b)

From the Newton’s first law, the total distance travelled by the car is expressed as:

$$\begin{gathered} s=1/2a{t}^2+ut \\ s=l+ut \end{gathered}$$

Here, l is the distance the car moved that is$75.5m$ , u is the initial velocity of the car that is$20m/s$ and t is the time taken by the car that is $15.86s$.

Substituting the values in the above equation, we get-

$$\begin{gathered} s=75.\text{5}m+20m/s^2×15.86s \\ s≈393m \end{gathered}$$

Thus, the distance that the car travelled during this time is$393m$ .

c)

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

$v=u+at$

Here, v is the final speed of the car, u is the initial speed of the car that is $20m/s$and t is the time taken by the car that is$15.86s$.

Substituting the values in the above equation, we get-

$$\begin{gathered} v=20m/s+\left({\text{0.6m/s}}^2\right)×\left(\text{15.86s}\right) \\ v=29.5\frac{m}{s} \end{gathered}$$

Thus, the final speed of the car is$29.5m/s$ .