Solve 1D Motion: Find Distance, Time, Velocity

AI Thread Summary
The discussion revolves around solving a physics problem involving two objects: one starting from rest with acceleration and the other moving at a constant velocity. Key calculations include determining the distance before they cross again, the time taken to do so, and the velocity of the accelerating object at that point. Participants clarify that the plotting of their positions should begin at the intersection where the traffic light is located, with one object represented by a straight line and the other by a parabolic curve. The equations provided are used to derive the necessary values for distance, time, and velocity. The conversation emphasizes the importance of correctly interpreting the initial conditions and plotting requirements for the problem.
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Homework Statement


at the instant the traffic light turns green, s starts from rest and accels at a_s. at the same time k passes s with constant velocity v_k.

given a_s, v_k,
find distance before the cross again
time it takes to do so
velocity of s at this point

plot the position as a function of time for both. take x=0 at the intersection <- no idea what last sentence means.

Homework Equations


v_s = a_s + t
d = 1/2 a_s * t^2
v_s^2 = 2a_s*d

d= v_k * t


The Attempt at a Solution



t = sqrt(2*v_k/a_s)
d = v_k * sqrt(2d/a_s)
v_s = a_s*sqrt(2d/a_s)


as for this plotting, does this mean I "start" the graph at their intersection? which one?
 
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as for this plotting, does this mean I "start" the graph at their intersection? which one?
............
Yes.
k with constant velocity-a straight line.
s with constant acceleration- a parabola.
 
oreosama said:
plot the position as a function of time for both. take x=0 at the intersection <- no idea what last sentence means.

...as for this plotting, does this mean I "start" the graph at their intersection? which one?

I think the "intersection" is the intersection of the streets where the traffic light is positioned. So, take x = 0 to be the location where k initially passes s.
 
wanted to be sure, thanks.
 
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