A particle is moving so that the acceleration at time t is:
a(t) = 24t^2 i - 12t j
The initial position and velocity are r(0) = i and v(0) = i - j respectively. Find r(t), the position at time t.
please help me with this problem.
Personally i like the vector notation better, so i will work in that and then convert it to i,j notation. You should be able to reproduce each step in i,j notation if you so desire
a(t) = <24t^2 , 12t>
=> v(t) = int(a(t)) = <8t^3 + C , 6t^2 + D>
But v(0) = <1,1>
so v(0) = <1,1> = < 0 + C , 0 + D>
=> C=D=1
so v(t) = <8t^3 + 1 , 6t^2 + 1>
r(t) = int(v(t)) = < 2t^4 + t + C , 2t^3 + t + D>
but r(0) = <1,0>
so r(0) = <1,0> = <0 + 0 + C , 0 + 0 + D>
=> C = 1, D = 0
so r(t) = <2t^4 + t + 1 , 2t^3 + t>
= (2t^4 + t + 1)i + (2t^3 + t)j
Hello, rcmango!
We will integrate twice, and use the initial conditions to evaluate the constants.A particle is moving so that the acceleration at time t is:
. . a(t) .= .24t²i - 12tj
The initial position and velocity are: .r(0) = i and v(0) = i - j, respectively.
Find r(t), the position at time t.
Velocity is the integral of the acceleration: .v(t) .= .∫ a(t) dt
. . v(t) .= .∫ (24t²i - 12tj) dt .= .8t³i - 6t²j + C1
Since v(0) = i - j: .8·0³i - 6·0²j + C1 .= .i - j . → . C1 .= .i - j
. . Hence: .v(t) .= .8t³i - 6t²j + i - j . → . v(t) .= .(8t³ - 1)i - (6t² + 1)j
Position is the integral of velocity: .r(t) .= .∫ v(t) dt
. . r(t) .= .∫ [(8t³ - 1)i - (6t² + 1)j] dt .= .(2t^4 - t)i - (2t³ + t)j + C2
Since r(0) = i: .(2·0^4 - 0)i - (2·0³ + 0)j + C2 .= .i . → . C2 = i
. . Hence: .r(t) .= .(2t^4 - t)i - (2t³ + t)j + i
Therefore: .r(t) .= .(2t^4 - t + 1)i - (2t³ + t)j
Hmmm, too slow once again . . .