Q67P - University Physics with Modern Physics

Question

The driver of a car wishes to pass a truck that is traveling at a constant speed of $\mathbf{20}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{/}\mathit{s}$ (about$\mathbf{41}\mathbf{mil}\mathbf{/}\mathbf{h}$ ). Initially, the car is also traveling at$\mathbf{20}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{/}\mathit{s}$ , and its front bumper is $24.0m$ behind the truck’s rear bumper. The car accelerates at a constant 0.600$m/s^2$, then pulls back into the truck’s lane when the rear of the car is $26.0m$ ahead of the front of the truck. The car is  long, and the truck is 21. 0m long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?

Step-by-Step Solution

Answer

a) Time required by the car to pass the truck is ,$15.86s$

 b) The distance that the car travelled during this time is 393m and 

c) The final speed of the car is 29.5m/s.

Step 1: Identification of the given data

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.

Step 2: Significance of Newton’s first law in evaluating the time, distance and the speed of the car

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.

Step 3: Determination of the time, distance and the speed 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-

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

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-

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

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-

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

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