position of a particle in frame s is given by r=(6t^2-4t)i-3t^3j+3k.in another frame s',its position of the same particle is given by r'=(6t^2+2t)i-(3t^3+8t)j+3k.what is the magnitude of relative velocity of s' with respect to s if r and r'are measured in metre and t in second.
I do not understand what you mean by a "frame". If is is "the same particle" in both "frames", why is the time dependence of the position vector different?
If you are talking about two different particles following different trajectories, then the problem makes sense. Just take the time derivative of the r' - r vector.
position of a particle in frame s is given by r=(6t^2-4t)i-3t^3j+3k.in another frame s',its position of the same particle is given by r'=(6t^2+2t)i-(3t^3+8t)j+3k.what is the magnitude of relative velocity of s' with respect to s if r and r'are measured in metre and t in second.
I know the question, which you have merely repeated. You have not responded to my comments and suggestions
To find the magnitude of the relative velocity of frame s' with respect to frame s, we need to calculate the difference between the velocities of the particle in each frame.
The velocity of a particle in frame s is obtained by taking the derivative of its position with respect to time (t):
v = dr/dt
For the position vector r = (6t^2-4t)i - 3t^3j + 3k, let's calculate the velocity vector in frame s:
v = d/dt((6t^2-4t)i - 3t^3j + 3k)
= (12t-4)i - (9t^2)j + 0k
= (12t-4)i - 9t^2j
Now, let's find the velocity vector in frame s' using the position vector r' = (6t^2+2t)i - (3t^3+8t)j + 3k:
v' = d/dt((6t^2+2t)i - (3t^3+8t)j + 3k)
= (12t+2)i - (9t^2+8)j + 0k
= (12t+2)i - (9t^2+8)j
Finally, we can find the relative velocity, v_rel, by subtracting the velocity of s from the velocity of s':
v_rel = v' - v
= [(12t+2)i - (9t^2+8)j] - [(12t-4)i - 9t^2j]
= 12t(i) - 12t(i) + 2(i) + 4(i) - 9t^2(j) - (-9t^2(j)) - 8(j)
= 6t(i) - 8(j)
The magnitude of the relative velocity, |v_rel|, is given by:
|v_rel| = sqrt((6t)^2 + (-8)^2)
= sqrt(36t^2 + 64)
= sqrt(36t^2 + 64)
Hence, the magnitude of the relative velocity of frame s' with respect to frame s is sqrt(36t^2 + 64).