Question

A particle position as a function of time t is given by r=(5.0t+6.0t^2)mi+(7.0-3.0t^3)mj. At t=5s find the magnitude and direction of the particles displecment vector delta(r) relative to the point r0=(0.0i+7.0j)m

Express your answer using two significant figures.

Answer 1

The magnitude of the particle's displacement vector at t=5s is 68 mi and the direction is approximately 158.2 degrees.

Step-by-step explanation:

To find the displacement vector at t=5s, we need to substitute t=5 into the given position function r=(5.0t+6.0t^2)mi+(7.0-3.0t^3)mj. This gives us r=(5.0(5)+6.0(5)^2)mi+(7.0-3.0(5)^3)mj. Evaluating this expression gives us r= 150mi - 218mi. Therefore, the displacement vector is delta(r) = 150mi - 218mi = -68mi.

The magnitude of the displacement vector is the absolute value of -68, which is 68 mi. The direction of the displacement vector can be determined by finding the angle it makes with the positive x-axis. Using the arctan function, we can find the angle to be approximately 158.2 degrees.