# Homework Help: Motion in 1D problem, it's not supposed to be this hard.

1. Aug 29, 2010

### Executioner

1. The problem statement, all variables and given/known data

To save fuel, some truck drivers try to maintain a constant speed when possible. A truck traveling at 91.0 km/hr approaches a car stopped at the red light. When the truck is 115.7 meters from the car the light turns green and the car immediately begins to accelerate at 2.90 m/s^2 to a final speed of 106.0 km/hr. How close does the truck come to the car assuming the truck does not slow down?

2. Relevant equations

I might be wrong but I've used this formula:
Vo = "V knot"
Xo = "X knot" (initial position)

X = (1/2)at^2 + Vot + Xo

3. The attempt at a solution

I convert the trucks velocity from 91 km/hr into 25.278 m/s, and I've converted the car's final velocity to 29.44 m/s. I tried to do the above equation for both vehicles and set them equal to each other so I'll know what time they meet. I was thinking this way and then when I find TIME, I'll plug it in back into one of the "X = (1/2) etc" equation to find the distance.

My professor doesn't show us any examples, all he did was just show us how we get the equation. Thanks guys for any help, I really appreciate it.

2. Aug 29, 2010

### ehild

The problem asks the minimal distance the truck approaches the car. If you find a time when they meet it means that the truck and the car will collide. Just think: The truck goes with uniform speed. The car accelerates from zero speed. Its speed increases, reaches that of the truck and gets even higher up to 106 km/h. Up to what speed will the distance decrease between them?

ehild

3. Aug 29, 2010

### novop

The truck is gaining on the car right up until the cars speed > the trucks speed. At what time does the cars speed equal the trucks speed? Where are the truck and car at this time?

4. Aug 29, 2010

### Executioner

Thanks guys, so at some point the car AND the truck both have the same velocity right?

5. Aug 29, 2010

### ehild

Yes, and you have to find that moment.

ehild

6. Aug 29, 2010

### Executioner

OK, so the car will hit the speed of 25.278 (which is the same like the truck) at some time T right? I find that T and then plug it back into the position equation for each of the vehicles right? After that subtract the two and I get how much farther (closest) they are apart?

7. Aug 29, 2010

### Executioner

Awesome! I got it! I knew it had something to do with time when they both has the same velocity! You just confirmed it and helped me out! Thank you so much guys!!!!! :)

8. Aug 29, 2010

### ehild

Congratulation! Good job!

ehild