A particle traces the path vector r = (t^3, t+3, t^2 -4t) where t represents time; it flies off on a tangent when t=1 and then moves in a straight line with constant speed. Find its position when t=4.
Do I just plug the 4 into the t's in vector r?
IT seems to me that the vector r is only valid to time=1, after that the particle is on a "tangent" at constant speed.
So find the position and velocity of the particle at time=1, then the velocity is constant for the next three seconds, find the new position. That is what I read into the problem.
To find the position of the particle when t=4, you can indeed substitute t=4 into the vector r = (t^3, t+3, t^2 -4t). However, it is important to note that at t=1, the particle flies off on a tangent, which means it will continue to move in a straight line with constant speed.
Let's break down the steps to find the position when t=4:
1. Start with the given vector equation: r = (t^3, t+3, t^2 - 4t)
2. Substitute t=4 into the equation. This means replacing t with 4 in all three components: r = (4^3, 4+3, 4^2 - 4*4)
3. Simplify each component: r = (64, 7, 16-16)
4. Calculate the final position: r = (64, 7, 0)
Therefore, the position of the particle when t=4 is (64, 7, 0).