Question

The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.0 m/s. Initially, the car is also traveling at a speed 19.0 m/s and its front bumper is a distance 23.1 m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.570 m/s^2, then pulls back into the truck's lane when the rear of the car is a distance 25.0 m ahead of the front of the truck. The car is of length 5.00 m and the truck is of length 21.3 m.
1. How much time is required for the car to pass the truck?
2. What distance does the car travel during this time?
3. What is the final speed of the car?

Answer 1

Answer:

16.16 s

381.5 m

28.21 m/s

Explanation:

Acceleration of the car = $0.57\ \text{m/s}^2$

The distance the car moves is $23.1+25+21.3+5=74.4\ \text{m}$

Initial position of the car is $23.1+21.3+5=49.4\ \text{m}$

Position of the truck is given by the equation

$x_r(t)=49.4+19t$

Position of the car is given by the equation

$x_c(t)=19t+\dfrac{1}{2}\times0.57t^2$

Difference in their positions is 25 m

$x_c(t)-x_r(t)=25\\\Rightarrow 19t+\dfrac{1}{2}\times0.57t^2-(49.4+19t)=25\\\Rightarrow 0.285t^2-74.4=0\\\Rightarrow t=\sqrt{\dfrac{74.4}{0.285}}\\\Rightarrow t=16.16\ \text{s}$

Time required for the car to pass the truck is 16.16 s

$x_c(16.16)=19\times 16.16+\dfrac{1}{2}\times0.57\times 16.16^2=381.5\ \text{m}$

The distance the car traveled during this time is 381.5 m

$v_c(16.16)=19+0.57\times 16.16=28.21\ \text{m}$

The final speed of the car is 28.21 m/s