Cars moving toward each other problem involving constant acceleration

[PatrickR.]

1. a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at xr = 0 and the green car is at xg = 240 m. If the red car has a constant velocity of 21.0 km/h, the cars pass each other at x = 47.5 m, and if it has a constant velocity of 44.0 km/h, they pass each other at x = 76.6 m. What are (a) the initial velocity (in km/h) and (b) the acceleration of the green car?

So I find that if xr= 47.5m, then xg= 192.5m and in the second case xr=76.6m and xg=163.4m

so i find time in the 2 cases. 47.5e-3 km / 21km/hr and 76.6e-3km / 44km/hr

= .00226hr and .001741 hr respectively , so instantaneous velocity for green in case 1 is

192.5e-3km / .00226 hr = 85.177km/hr and 163.4e-3km / .001741 = 93.854km/hr


I use v= v0 + at to find that a= -16718km/hr^2 = -1.29m/s^2

I use the same equation again v= v0 + at to find v0 at the very beginning of the road.

93.854= v0 + (-16718.7km/hr^2)(.001741hr)

v0= 122.96km/hr

However, these answers are not correct so I would like to know what I am doing wrong. Thankyou

 

[loonychune]

You're going to have to take a different approach to this problem.

What you've done in finding the 'instantaneous velocity' is actually found the average velocity over the time it takes from x_g to the point of crossing the red car -- but the car is accelerating!

 

[PatrickR.]

You're absolutely right hmm... now to find a different approach.

 

[loonychune]

Try setting x = 0 at the xg = 240 point... then in the following equation s will be the distances you worked out to crossing but will be negative

$$s = ut + \frac{1}{2}at^2$$

You've worked out s and you know how long it takes to get to s, i.e. t, in two cases..

Hence you'll just have two simultaneous equations to solve