If a particle moves with velocity V PA = (2.0 m/s) + (4.0 m/s) relative to frame A, what is the velocity V AP of the frame relative to the particle?
a. -(2.0 m/s) + (4.0 m/s)
b. -(2.0 m/s) - (4.0 m/s)
c. (2.0 m/s) + (4.0 m/s)
I'm going to make the assumption that this is in vector unit notation and that this is a two dimensional problem.
If particle P is moving 2.0m/s(i) + 4m/s(j) with respect to particle A then particle A should appear to be moving -2.0m/s(i) - 4.0m/s(j). This can be imagined by assuming that one particle is stationary and finding the motion of the other particle relative to that particle.
Particles in motion appear to be stationary in their reference frame so everything else seems to move relative to them. Therefore if particle A sees particle P moving at a magnitude of 4.4m/s relative to itself. P may view the same system such that particle particle A, relative to particle P, moves at -4.4m/s magnitude, or choice B.
The velocity V AP of the frame relative to the particle is equal in magnitude but opposite in direction to the velocity V PA of the particle relative to the frame. Therefore, the correct answer is:
b. -(2.0 m/s) - (4.0 m/s)
To find the velocity V AP of the frame relative to the particle, we need to use the concept of relative velocity. The velocity of frame A relative to the particle is equal in magnitude but opposite in direction to the velocity of the particle relative to frame A.
Given that the velocity V PA of the particle relative to frame A is (2.0 m/s) + (4.0 m/s), we can determine the velocity V AP of the frame relative to the particle as follows:
V AP = -V PA
In other words, to find the velocity V AP, we need to change the sign of each component of V PA. This means that the correct answer is:
b. -(2.0 m/s) - (4.0 m/s)
So, the velocity of frame A relative to the particle is -(2.0 m/s) - (4.0 m/s).