Consider the motion of a particle that experiences a variable acceleration given

Question

Consider the motion of a particle that experiences a variable acceleration given by a_x = a_0 + bt, where 3_0 and b are constants and x = x_0 and v_x = v_0 at t = 0. (Use the following as necessary: a_0, v_0, x_0, b, and t.) Find the instantaneous velocity as a function of time. Find the position as a function of time. Find the average velocity for the time interval with an initial time of zero and arbitrary final time t. Compare the average of the initial and final velocities to your answer to Part (c). Are these two averages equal?

Explanation / Answer

5.

ax = ao + bt

d Vx/dt = ao + bt

Vx = ao t + (0.5) b t2 + C

At t = 0

Vx = Vo = C

so equation becomes

Vx = ao t + (0.5) b t2 + Vo

b)

dx/dt = ao t + (0.5) b t2 + Vo

X = (0.5) ao t2 + 0.17 b t3 + Vo t + D

at t = 0

X = Xo = D

so X = (0.5) ao t2 + 0.17 b t3 + Vo t + Xo

c)

at t = 0 , initial Position = Xo

at t = t , final Position = (0.5) ao t2 + 0.17 b t3 + Vo t + Xo

Displacement = final position - initial position = (0.5) ao t2 + 0.17 b t3 + Vo t + Xo - Xo = (0.5) ao t2 + 0.17 b t3 + Vo t

time = t

average velocity = displacement / time = ((0.5) ao t2 + 0.17 b t3 + Vo t ) / t = (0.5) ao t + 0.17 b t2 + Vo

d) average = Vo + Vx/2 = (ao t + (0.5) b t2 + Vo + Vo) / 2

those are not equal

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