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.

Printable View

- February 27th 2007, 05:32 PMrcmangomoving particle
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. - February 27th 2007, 06:02 PMJhevon
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 - February 27th 2007, 06:26 PMSoroban
Hello, rcmango!

Quote:

A particle is moving so that the acceleration at time t is:

. .**a**(t) .= .24t²**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*.

Velocity is the integral of the acceleration: .**v**(t) .= .∫**a**(t) dt

. .**v**(t) .= .∫ (24t²**i**- 12t**j**) dt .= .8t³**i**- 6t²**j**+ C1

Since**v**(0) =**i**-**j**: .8·0³**i**- 6·0²**j**+ C1 .= .**i**-**j**. → . C1 .= .**i**-**j**

. . Hence: .**v**(t) .= .8t³**i**- 6t²**j**+**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 . . . - February 27th 2007, 06:41 PMJhevon
- February 27th 2007, 09:27 PMrcmango
awesome explanations, thanks for all the work!