Two cars at same speed, one passes the other

[MissJewels]

Homework Statement

This question has gotten me stuck. I translated it from french to english so pardon that... I just get lost when you include two objects in physics.

A car and a truck move initially in the same direction at 20 m / s, the truck with 38m ahead. The car accelerates at a constant rate of 2m / s ^ 2 exceeds the truck, and falls back into the right lane when it is 11m in front of the truck.
a) How far has the truck traveled in that time?
b) For what value of constant acceleration of the truck it and the car will they have a velocity that differs by only 5 m / s when they intersect? It is assumed that the truck begins to accelerate when a lead of 38m on the car.

Homework Equations

d= Vi*t + (0,5)at^2
d= distance of the car

The Attempt at a Solution

I tried
But I keep getting negative answers! Someone give me a boost or help me start it! Please!

...
answers:
a) 140m
b) a(car) > 1,67 > s^2

 

[DukeLuke]

I get the same answer for part a. For part b you know that when they intersect the position (x value) for both the car and truck are equal (xcar=xtruck) where you can apply this equation for both the car and truck

[tex]
x = x_0 +v_0t+\frac{1}{2}at^2
[/tex]

You also know that at this point (same t) the velocities differ by 5m/s (vcar-vtruck = 5m/s) where you can apply this equation for both the car and truck

[tex]
v = v_0 +at
[/tex]

This should give you two equations and two unknowns (t and a).

 

[MissJewels]

Thanks! But, I am still stuck on a) though... I can't see how i can mesure the distance pursued bby the truck..

 

[DukeLuke]

I solved for the time it took for the car to get 11m in front of the truck and then put that into the equation for the truck (not accelerating xtruck = xo + vot). I also got 140m, so unless I'm wrong you have the right answer.

 

[MissJewels]

I didnt find those answers, those are the correct answers provided by my book.

 

[MissJewels]

I just plugged in these in the equation for a) and found the time to catch up with the truck, i think...:
d = Vi*t + (0,5)at^2
38 = 20t + (0,5)2t^2

but i don't think that's right, because the truck is always moving forward, not staying at 38m...

 

[MissJewels]

I got some help from a tutor and Seems I've been a tad blind, thank you for helping, means a lot!