A "rocket car" is launched along a long straight track att = 0 s. It moves with constant acceleration a1 = 2.9m/s2 .At t = 2.8s , a second car is launched with constant acceleration a2 = 7.0m/s2 .
At what time does the second car catch up with the first one?
How far down the track do they meet?
[itex]v_f^2 = v_i^2 + 2a(Δx)[/itex]
[itex]v_f = v_i + at[/itex]
[itex]x_f = x_i + v_it + 1/2at^2[/itex]
The Attempt at a Solution
[itex]x_f = (1/2)(2.9m/s^2)(2.8s)^2 = 11.368m[/itex]
Car 1's X initial is 11.368m when the second car is launched.
Equation for X final of car 1:
[itex]x_f = (0.5)(2.9m/s^2)(t)^2 + (8.12m/s)(t) + 11.368m[/itex]
Equation for X final of car 2:
[itex]x_f = (0.5)(7.0m/s^2)(t)^2[/itex]
Setting X final equal to each other (when second car meets up with first one)
[itex](0.5)(7.0m/s^2)(t)^2 = (0.5)(2.9m/s^2)(t)^2 + (8.12m/s)(t) + 11.368m[/itex]
[itex] t = 5.06s[/itex]
It turns out 5.06s is incorrect. Not sure what I am doing wrong here. Just figured out how to use the fancy math notation! My post is a well-crafted scientific paper.