Question

The position of a particular particle as a function of time is given by r=(9.60ti^+8.85j^​−1.00t2k^)m, where t is in seconds. Determine the particle's velocity as a function of time. Express your answer in terms of the unit vectors i^,j^​, and k^. Part B Determine the particle's acceleration as a function of time. Express your answer in terms of the unit vectors i^,j^​, and k^.

Answer 1

Answer: the acceleration of the particle as a function of time is a = -2.00k^ m/s².

Explanation:

To find the velocity of the particle as a function of time, we need to take the derivative of the position vector with respect to time.

Given: r = (9.60ti^ + 8.85j^ - 1.00t^2k^) m

Taking the derivative with respect to time (t):

v = dr/dt

Differentiating each component of the position vector:

v = (d(9.60t)/dt)i^ + (d(8.85)/dt)j^ + (d(-1.00t^2)/dt)k^

v = 9.60i^ + 0j^ - 2.00t k^

Therefore, the velocity of the particle as a function of time is v = (9.60i^ - 2.00tk^) m/s.

For Part B, to determine the particle's acceleration as a function of time, we need to take the derivative of the velocity vector with respect to time.

Given: v = (9.60i^ - 2.00tk^) m/s

Taking the derivative with respect to time (t):

a = dv/dt

Differentiating each component of the velocity vector:

a = (d(9.60)/dt)i^ + (d(-2.00t)/dt)k^

a = 0i^ - 2.00k^