Question

Two cars pass one another while travelling on
a city street. Using the velocity-time graph
below, draw the corresponding position-time
graph and determine when and where the two
cars pass one another. Assume both cars start at
position 0.0 m [N] and time 0.0 s.

Answer 1

@My15girl
where's the graph?

Answer 2

Darn, okay I'll try to get a picture of it.

Answer 3

\[d = vt. \space Set \space d_1 = d_2 : v_1t = v_2t\]
One of the cars is accelerating, so you can either use v=at for its velocity, or since it is accelerating linearly, use its average velocity during the time interval.

Answer 4

I don't understand how you find it. For d1 the area under the vt graph is 120m and for d2 the the area under the graph is 126. So idk how do I solve this?

Answer 5

I've tried several ways and I still can't solve this. :(

Answer 6

What you need are the position functions.
One has d = 12t for constant velocity, the other has d = 1.5t^2, t≤6 + 18(t-6), t > 6.

Answer 7

They definitely don't meet before 6 seconds, so set d1 = d2 : 12t = 54 + 18(t-6) and solve for t.

Answer 8

okay. I'll try doing that.

Answer 9

Thank you

Answer 10

Does it make sense?

Answer 11

Yeah. It actually makes a lot more sense now. My physics teacher sometimes confuses me..so it makes so much more sense, however, what did you mean by d = 1.5t^2, t≤6. Did you just mean that the position would be 1.5^2 and the time would be bigger than 6

Answer 12

No. The velocity function for the one that starts at 0 m/s is a piece-wise function. It is linear with constant acceleration until t=6s, then is constant after.

Answer 13

The slope of the velocity curve is the acceleration, so you have to use .5at^2 for displacement, after t=6, the acceleration is zero, so the displacement function is vt for v=18 m/s and t>6s