Motion Along A Straight Line - HELP

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SUMMARY

The discussion focuses on solving a physics problem involving motion along a straight line, specifically the scenario where a car accelerates from rest at 3.15 m/s² while a truck moves at a constant speed of 18.0 m/s. To determine how far the car overtakes the truck and its speed at that moment, participants suggest using the equations of motion: x(t) = At² + Bt + C for position and v(t) = 2At + B for velocity. The constants A, B, and C must be defined based on initial conditions to find the intersection point of both vehicles' paths.

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At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.15 . At the same instant a truck, traveling with a constant speed of 18.0 , overtakes and passes the car.

Part A: How far beyond its starting point does the car overtake the truck?
Part B: How fast is the car traveling when it overtakes the truck?



Vo of the car is zero and the acc. = 3.15
Vo of the truck is 18 and the acc. is zero,
i need atleast a hint,i don't get part A,i don't have enough information:S But i think part B would be easy once i know part A
 
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Basically, you need to find the equations of motion for both vehicles. It means the dependence of coordinate upon time. In your case it has the form

[tex]x(t) = At^2 + Bt + C[/tex]

and velocity

[tex]v(t) = 2At + B[/tex]

Just fix the constants A, B, and C to meet the boundary conditions. Once you know the constants for both vehicles, you can equate their positions to get the time(s) when they meet and then calculate the point of meeting, as well as any other quantity of interest.
 
I'm sorry but I don't get it, from where do i bring A,B and C?

I think the only equations we can use are:

[tex]V = Vo + at[/tex]
[tex]V^2 = Vo^2 + 2aX[/tex]
[tex]X = volt + .5 at^2[/tex]
[tex]X = Vt - .5 at^2[/tex]
[tex]X = .5 (Vo+V) t[/tex]

V being final velocity, Vo initial velocity, t time, a acceleration,X displacement.
 

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