# Homework Help: A car and a truck

1. Aug 26, 2007

### azila

1. The problem statement, all variables and given/known data
Thr driver of a car wishes to pass a truck that is traveling at a constant speed of 19.3 m/s. Initially, the car is also traveling at a speed 19.3 m/s and its front bumper is a distance 23.9 m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.600 m/s^2. Then pulls back into the truck's lane when the reat of the car is a distance 25.8 m ahead of the front of the truck. The car is of length 4.5 m and the truck is of length 21.4 m .

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

2. Relevant equations

I was using the equation where xcar = Xo + Voxt + (1/2)at
and the xtruck = Xo + Voxt + (1/2)at. The problem is that I know that if I were asking how much time does it take for the car to be at the same position as the truck I would set the two equations equal to each other, but I am asking how long does it take to overtake the truck. I even tried, adding the distances but I am getting more and more confused.

3. The attempt at a solution

Vintially of Car = 19.3 m/s Vinitially of Truck = 19.3 m/s
acceleration of car: .600 m/s^2 Acceleration of Truck = 0
X Distance of Car = 0 X distance of Truck = 23.9

Now, would I set them equal to each other to get the time that it takes the car to overtake the truck or should I add distances? I am really confused. Please HELP ME!!!

2. Aug 26, 2007

### Mindscrape

What are you setting equal to each other? What are the two displacement equations of the car and the truck? Split the problem into pieces if you need to.

3. Aug 26, 2007

### azila

By saying that I am going to set them equal to each other, I meant would i need to set the equations equal to each other like this:

xcar = Xo + Vot + (1/2)at xtruck = Xo + Vot + (1/2)at
xcar = 0 + 19.3t + (1/2)(.600)t xtruck = 23.9 + 19.3t + (1/2)(0)t
xcar = 19.3t + .3t x truck = 23.9 + 19.3t

19.3t + .3t = 23.9 + 19.3t
.3t = 23.9
t = 79.7 sec?????

The answer though is not right; I don't know what to do. Am I even in the right path or direction of solving. I am working it like a problem in the textbook that shows how to work it. Thanks for replying but I am stuck.

4. Aug 26, 2007

### mgb_phys

How far does the car have to go?
s(car) = rear distance + length of truck + lenght of car + distance in front + extra distance truck moved
s(car) = 23.9 + 21.4 + 25.8 + 4.5 + v(truck)t
s(car) = 75.6 + v(truck)*t = 75.6 + 19.3t

And as you said
s(car) = ut + 1/2 at^2 = 19.3t + 0.5 * 0.6 * t^2
Set these equal to each other and solve for t

(ps always check my numbers I might have made an arithmetic error!)

5. Aug 26, 2007

### azila

Thank You So Much!!!!! You Don't Know How Much You Have Helped Me!! Thanks Again!!!