One-Dimensional Kinematics problem

[tchronos]

Hi, i have some questions for the following problems.

A car accelerate at -1.9 m/s2. Traveling at a constant velocity of 32 m/s, this car comes up behind a car traveling at a constant velocity of11 m/s.

a) How close to the slower car can the driver of the faster car come before applying his brakes and still avoid a collision?

b) At what time does the inevitable collision of the two cars occur?

c) How far beyond its position at t = 0 does the slower car get before it is hit?

so the |a1| = 1.9 m/s2
v1 = 32 m/s
v2 = 11 m/s.

Since it accelerates in a negative direction then it is deccelerating at a constant rate. For every second the velocity with decrease by 1.9 until it crashes with the second car with has a constant velocity. Now I don't understand what the question is asking for part a)

Can anyone help? thanks

 

[dextercioby]

Write the equations of motion for each car and then impose the conditions that fulfiled for the crash to happen.

Daniel.

 

[tchronos]

hi thanks for the reply.
I found that the two equations for the position.

X1 = V1t + .5 at^2
X2 = D + V2t

If i set these two equal, I should find the point of collision but there are two variables and they don't give me D or T. If they have given me D then the problem will be a lot easier.

On the first problem, I want to find the D but they did not provide me with a time.

 

[HuGaPdBeAr]

Try using the velocity equation for car 1 to find t.

 

[69Mustang]

Better yet, use concept of relative velocity and relative acceleration. Set the final relative position at 0 and solve for initial position.