A particle starts with speed and moves in a straight line. The particle is subject to a retardation which is initially and which increases uniformly with the distance moved, having a value of when the particle has moved . Given that the particle has speed and has moved a distance of , show that, while the particle is in motion
My Problem with this question:
I can't derive it. Help please? Thanks in advance.
Also, a follow on question is:
Hence, or otherwise, calculate the distance moved by the particle in coming to rest.
I worked it out to get two answers, and . I accepted which is correct but realised that they are asking for distance so doesn't that mean that it could also be ?
I'll take moving right as the positive direction and moving left as the negative direction. There are two cases to consider: v = 20 m/s (the particle moves to the right with an intial speed of 20 m/s) and v = -20 m/s (the particle moves to the left with an initial speed of 20 m/s). You have .
1. v = 20 m/s. Then , x > 0.
Draw a graph of v versus x. Note that it starts at v = 20 and x = 0 (corresponding to t = 0) and x > 0. It's very clear that when v = 0, x = 20. That is, the distance travelled is 20 m.
2. v = -20 m/s. Then , x < 0.
Draw a graph of v versus x. Note that it starts at v = -20 and x = 0 (corresponding to t = 0) and x < 0. It's very clear that when v = 0, x = -40. That is, the distance travelled is 40 m.
The origin of your two solutions should now be clear. Clearly the book has assumed v = 20 m/s .......
By the way, the graph of is an ellipse.