You are in the lab, monitoring the trajectory of a particle. You measure the par

Question

You are in the lab, monitoring the trajectory of a particle. You measure the particle's position as a function of time and find that it can be described as: r = (2.9x + 1.5t^2 y + 0.3t z) m a. Determine the velocity of the particle as a function of time. b. Determine the acceleration of the particle as a function of time. c. what is the magnitude of the instantaneous velocity of the particle at t = 2 s? d. What is the average velocity of the particle between t = 1 and t = 4 s? e. what is the average acceleration of the particle between t = 1 and t = 4 s?

Explanation / Answer

Particles position,

                r= (2.9 x + 1.5 t^2 y +0.3 t z ) m

Part (a)

Velocity of particle is

                V=dr/dt = d/dt (2.9 x + 1.5 t^2 y +0.3 t z ) = 0 x + 3.0 t y +0.3 z

Part (b)

Acceleration of particle is

                a=dv/dt = d/dt (0 x + 3.0 t y +0.3 z ) = 0 x + 3.0 y + 0 z

Part (c)

Instantaneous Velocity of particle at t=2 s is

                V= 0 x + 3.0 t y +0.3 z   = 0 x + 3.0 (2) y +0.3 z = 0 x + 6.0 y +0.3 z

Magnitude of v is

                V= sqrroot(Vx^2 +Vy^2 +Vz^2) = sqrroot (0+36+0.09) =6.008 m/s

Part (d)

Average velocity is

                Vavg = r(t2) – r(t1) /t2-t1   = r(t=4) – r(t=1) /4-1

= (2.9 x + 1.5 (4)^2 y +0.3 (4) z ) - (2.9 x + 1.5 (1)^2 y +0.3 (1) z ) / 3

= 0 x + 7.5 y +0.3 z

Part (e)

Average acceleration

                a= v(t2) – v(t1) /t2-t1   = v(t=4) – v(t=1) /4-1

                   =   (0 x + 3.0 (4) y +0.3 z ) – (0 x + 3.0 (1) y +0.3 z ) /3

                    = 0 x + 3.0 y +0 z

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