Question

The position of a particle is →r(t)=(3.0t^2ˆi+5.0ˆj−6.0tˆk)m. (a) Determine its velocity and acceleration as functions of time. (b) What are its velocity and acceleration at time t = 0?

a) (i) v(t)=6.0tˆi m/s, a(t)=-6.0ˆk m/s²
(ii) v(t)=6.0tˆi m/s, a(t)=6.0ˆk m/s²
(iii) v(t)=6.0ˆi m/s, a(t)=6.0tˆk m/s²
(iv) v(t)=-6.0ˆi m/s, a(t)=-6.0ˆk m/s²

b) (i) v=0 m/s, a=0 m/s²
(ii) v=3.0 m/s, a=-6.0 m/s²
(iii) v=0 m/s, a=6.0 m/s²
(iv) v=3.0 m/s, a=6.0 m/s²

Answer 1

The velocity function of the particle is v(t) = 6.0tˆi - 6.0ˆk m/s and the acceleration function is a(t) = 6.0ˆi m/s^2. At time t = 0, the velocity is 0 m/s and the acceleration is 6.0ˆi m/s^2.

Step-by-step explanation:

To find the velocity and acceleration as functions of time for a particle with position function →r(t)=(3.0t^2ˆi+5.0ˆj−6.0tˆk)m, we differentiate the position function with respect to time.

Velocity

Velocity, v(t), is the first derivative of position:

v(t) = d(→r(t))/dt = 6.0tˆi - 6.0ˆk m/s

Acceleration

Acceleration, a(t), is the derivative of velocity:

a(t) = dv(t)/dt = 6.0ˆi m/s2

At time t = 0

To find the velocity and acceleration at t = 0, we substitute t = 0 into the velocity and acceleration functions:

v(0) = 0 m/s (since the term with t becomes 0)

a(0) = 6.0ˆi m/s2 (since acceleration is constant)