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
Average velocity:
Velocity at time
Velocity at
Explanation:
I'll assume you mean the average velocity from
We're given the components of the particle's acceleration, and asked to find the average velocity over the first
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:
Second integration:
The initial position is assumed to be at the origin, so let's plug in
We can then split the average velocity equation into components:
Using these components, we can find the magnitude of the average velocity vector:
(Here's the instantaneous velocity section) .
To find the instantaneous velocities at
#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"#