# Homework Help: Acceleration and velocity question-easy-need help

1. Sep 17, 2011

### vicsic

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

A car is waiting at a stoplight. Just as the light turns green and the car starts to accelerate, a truck passes at a constant velocity in the next lane. The truck passes at a velocity 25.0 m/s while the car accelerates 5.00 m/s^2 both in the same direction,
Calculate the distance travelled by the car before it overtakes the truck?
Car's speed as it overtakes the truck?

2. Relevant equations

I am a little confused on this one.

3. The attempt at a solution

I think you have to use the formula vf^2=Vi^2 +2ad but am not sure. Do you combine 2 formulas?

any help would be appreciated

2. Sep 17, 2011

### USN2ENG

.............................

3. Sep 17, 2011

### vicsic

When I solve for t it become zero, therefore the distance becomes zero which is impossible. any other suggestions

4. Sep 17, 2011

### Jokerhelper

You should probably show how you solved for time, because you clearly made an error at some point.

5. Sep 17, 2011

### USN2ENG

Recheck your equation. This is how it goes:

X = Xi + Vi(t) + (.5)Ax(t^2)
Xc = 0 + 0(t) + (.5)(5.00)(t^2) = 2.5t^2
XT = 0 + 25.0(t) + (.5)(0)(t^2) = 25.0t

SO....2.5t^2 = 25.0t.......THEN.....t=10.

6. Sep 17, 2011

### vicsic

car 1-

vi=0 m/s
a=5.00 m/s^2

Truck

v=25.0 m/s

assuming the distances are the same at the beginning
distance would equal zero as they are stopped

using the formula d=Vi(t)+1/2at^2

0=0t+1/2(5.00)t^2
0=1/2(5.00)t^2
0=5.00t^2
0=t^2
t=0

plug that back into the formula

D=Vf(t)-1/2at^2
D=25(t)-1/2(5.00)t^2
D=0

this is my work for this and it is not getting me the right answer which in the back of the book says 250m for A and 50 m/s for B

7. Sep 17, 2011

### USN2ENG

They are definitely not stopped considering the truck has a constant velocity.

The reason we set the distance equal to one another (NOT TO 0, because we dont know the distance) is that we are finding WHEN the car meets up with the trucks position. By solving for t, as I did above, you do that.

So if you look at my equation above and then plug in t=10 to either X function.....you get exactly 250m.

8. Sep 17, 2011

### USN2ENG

Also, the reason you were getting t=0 is that you were solving for when the cars were at distance 0...which is t = 0.

9. Sep 17, 2011

### vicsic

I am still a bit confused with the x's u are using in your formula. We were given the formula

d=vi(t) + 1/2at^2

how did u get all of those x's into it and 0's for the x's

thanks for all the help!

10. Sep 17, 2011

### USN2ENG

So the whole equation is:

d(the position of x from Xi aka the distance) = Xi(Initial position of x which is 0 and why they probably left it out) + Vi(t)(which is initial velocity, 0 for the car cause it is at rest in the beginning, and 25 is given for the truck) + 1/2Axt^2(I put the x because it just means Acceleration in the x direction, but you probably haven't went into acceleration in the y direction yet so that is why it is tripping you up. Just think that Ax is equal to what you have for a, Also, A is 0 for the truck because the velocity is constant, and 5.0 is given for the car.)

That should cover everything, hope that helps.

11. Sep 18, 2011

### vicsic

thanks i got the answer!

:)