Constant Acc with two particles

[pappaloo]

Im having a problem with this question, which is probably rather simple but I am making a lot harder than it needs to be.

A high-speed train is traveling at 44.72m/s and it sees a sidecar a distance of 676m ahead. The sidecar is traveling with a constant velocity or 8.056m/s.

The question asks what the magnitude of the resulting constant deceleration must be if a collision is to be just avoided.

I took two of the derived C.A. equations and in each one had two unknowns which were x(final) and t.

I then equated the two equations and solved the linear system for t at which I arrived with an answer of 47.26s

From there it is quite easy to figure out the rest of the question, but I am not sure if I am doing this question the best way or even correctly for that matter.

Any suggestions or comments would be greatly appreciated

 

[Doc Al]

I'm not sure exactly what you did, but here are two ways to approach it:
(1) Write the postions of each (sidecar and train) as a function of time. Set them equal, and solve for t. (This is what you tried, I believe.) Of course, the acceleration is a variable. Since the equation is a quadratic, you can find a constraint that will tell you the minimum acceleration that just avoids the collision.
(2) Look at the motion from the frame of the sidecar and solve it directly with a kinematic equation.

And welcome to PF!

 

[M98Ranger]

.

It seems like you are on the right track with your approach to this problem. You have correctly identified the two unknowns (x(final) and t) and used two equations to solve for them. However, there is a simpler way to approach this problem using the concept of relative velocity.

First, we can calculate the relative velocity between the train and the sidecar. This can be done by subtracting the sidecar's velocity from the train's velocity, since they are moving in the same direction. This gives us a relative velocity of 36.664m/s.

Next, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. We know the initial velocity (36.664m/s) and the final velocity (0m/s, since we want the train to come to a stop just before colliding with the sidecar). We also know the distance between the train and the sidecar (676m). Plugging these values into the formula, we get:

0 = 36.664 + a(47.26)

Solving for a, we get a deceleration of -0.775m/s^2. This means that the train needs to decelerate at a rate of 0.775m/s^2 in order to avoid a collision with the sidecar.

In summary, your approach was correct, but using the concept of relative velocity can simplify the problem and give you a quicker solution. Keep up the good work!