Two cars meeting - acceleration?

[Caitlin.Lolz.]

A car is traveling at a constant speed of 30 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 2.8 km away?

I have tried this problem many different ways and keep on coming up with .161 m/s^2. I made sure to convert the km to m, but the answer is still wrong!

If anyone can help me see what I am doing wrong, that would be great!

 

[alphysicist]

Hi Caitlin.Lolz.,

A car is traveling at a constant speed of 30 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 2.8 km away?

I have tried this problem many different ways and keep on coming up with .161 m/s^2. I made sure to convert the km to m, but the answer is still wrong!

If anyone can help me see what I am doing wrong, that would be great!

How did you come up with that answer? What was your procedure, what equations did you use, and what numbers did you use in the equations?

 

[dynamicsolo]

If you want a helper to see what you are doing wrong, it's a good idea to post your work.

How long did you find that it takes for the two cars to meet up? What is the kinematic equation you set up to describe the distance covered by the car starting at rest at the ramp?

 

[Caitlin.Lolz.]

the work I did was:

Vo = 0 m/s
d = 2.8 km (or 2800 m)
V = 30 m/s

d=.5(Vo+V)t
2800 = .5 (0 + 30) t
t = 186.67 seconds

d= Vo + .5 at^2
2800 = 0 + .5 (186.67)^2a
a = .161 m/s^2

 

[LowlyPion]

the work I did was:

Vo = 0 m/s
d = 2.8 km (or 2800 m)
V = 30 m/s

d=.5(Vo+V)t
2800 = .5 (0 + 30) t
t = 186.67 seconds

d= Vo + .5 at^2
2800 = 0 + .5 (186.67)^2a
a = .161 m/s^2

If the car starting from rest is going to catch the moving car by the next exit, then how much time does it have to do that?

Isn't it the distance to the next exit divided by the constant speed of the moving car and not the average speed of the car accelerating from rest?

Or put another way the average speed of the car starting at rest must be the speed of the car initially in motion. It must average the same speed (30m/s). To do that it's final speed will need to be 60m/s, not the 30m/s that you used.