What is the velocity of a particle for #t=0 to t=10# whos acceleration is #veca =3t^2 hati+5t hatj-(8t^3+400)hatk#?

1 Answer
Nathan L.
Jun 13, 2017

Average velocity: #6.01 xx 10^3# #"m/s"#

Velocity at time #t = 0# #"s"#: #0# #"m/s"#

Velocity at #t = 10# #"s"#: #2.40 xx 10^4# #"m/s"#

Explanation:

I'll assume you mean the average velocity from #t = 0# to #t = 10# #"s"#.

We're given the components of the particle's acceleration, and asked to find the average velocity over the first #10# seconds of its motion:

#vecv_"av" = (Deltavecr)/(10"s")#

where

  • #v_"av"# is the magnitude of the average velocity, and

  • #Deltar# is the change in postion of the object (from #0# #"s"# to #10# #"s"#).

We must therefore find the position of the object at these two times.

We have to derive a position equation from this acceleration equation, by integrating it two times:

First Integration:

#vecv = (t^3)hati + (5/2t^2)hatj - (2t^4 + 400t)hatk# (velocity)

Second integration:

#vecr = (1/4t^4)hati + (5/6t^3)hatj - (2/5t^5 + 200t^2)hatk# (position)

The initial position is assumed to be at the origin, so let's plug in #10# for #t# in the position equation:

#vecr = (2500)hati + (2500/3)hatk - (60000)hatk#

We can then split the average velocity equation into components:

#v_"av-x" = (Deltax)/(10"s") = (2500"m")/(10"s") = color(red)(250# #color(red)("m/s"#

#v_"av-y" = (Deltay)/(10"s") = (2500/3"m")/(10"s") = color(blue)(250/3# #color(blue)("m/s"#

#v_"av-z" = (Deltaz)/(10"s") = (-60000"m")/(10"s") = color(green)(-6000# #color(green)("m/s"#

Using these components, we can find the magnitude of the average velocity vector:

#v_"av" = sqrt((v_"av-x")^2 + (v_"av-y")^2 + (v_"av-z")^2)#

#= sqrt((250"m/s")^2 + (250/3"m/s")^2 + (-6000"m/s")^2)#

#= color(purple)(6.01 xx 10^3# #color(purple)("m/s"#

(Here's the instantaneous velocity section) .

To find the instantaneous velocities at #t = 0# and #t = 10# #"s"#, let's first plug in these times into the previously integrated velocity equation:

  • #t = 0# #"s"#

#vecv = ((0"s")^3)hati + (5/2(0"s")^2)hatj - (2(0"s")^4 + 400(0"s"))hatk#

#= color(red)(0# #color(red)("m/s"#

  • #t = 10# #"s"#

#vecv = ((10"s")^3)hati + (5/2(10"s")^2)hatj - (2(10"s")^4 + 400(10"s"))hatk#

#= (1000"m/s")hati + (250"m/s")hatj - (24000"m/s")hatk#

The magnitude of this velocity is thus

#v(10"s") = sqrt((1000"m/s")^2 + (250"m/s")^2 + (-24000"m/s")^2)#

#= color(blue)(2.40 xx 10^4# #color(blue)("m/s"#

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