Question

Question:

the particle is moving with acceleration a= 3ti + 2j at the time t seconds.
given that at time t= 0 the particle is at 4i + 2j and has the velocity I -4j ,find:
(a) The velocity of the particle at time t.
(b) position vector of the particle at time t.
(c) The distance of particle from the origin when t = 3.​

Answer 1

Use the fundamental theorem of calculus to find the velocity and position functions for the particle. We're given

a (t ) = 3t i + 2 j

v (0) = i - 4 j

x (0) = 4 i + 2 j

Then the velocity at time t is

`\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t \mathbf a(u)\,\mathrm du`

`\mathbf v(t)=(\mathbf i-4\,\mathbf j)+\displaystyle\int_0^t(3u\,\mathbf i+2\,\mathbf j)\,\mathrm du`

`\mathbf v(t)=(\mathbf i-4\,\mathbf j)+\left(\frac32 u^2\,\mathbf i+2u\,\mathbf j\right)\bigg|_0^t`

`\mathbf v(t)=(\mathbf i-4\,\mathbf j)+\left(\frac32 t^2\,\mathbf i+2t\,\mathbf j\right)`

`\mathbf v(t)=\left(\frac32 t^2+1\right)\,\mathbf i+(2t-4)\,\mathbf j`

and the position at time t is

`\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du`

`\mathbf x(t)=(4\,\mathbf i+2\,\mathbf j)+\displaystyle\int_0^t\left(\left(\frac32 u^2+1\right)\,\mathbf i+(2u-4)\,\mathbf j\right)\,\mathrm du`

`\mathbf x(t)=(4\,\mathbf i+2\,\mathbf j)+\left(\left(\frac12 u^3+u\right)\,\mathbf i+(u^2-4u)\,\mathbf j\right)\bigg|_0^t`

`\mathbf x(t)=(4\,\mathbf i+2\,\mathbf j)+\left(\frac12 t^3+t\right)\,\mathbf i+(t^2-4t)\,\mathbf j`

`\mathbf x(t)=\left(\frac12 t^3+t+4\right)\,\mathbf i+(t^2-4t+2)\,\mathbf j`

(a) As shown above,

v (t ) = (3/2 t ² + 1) i + (2t - 4) j

(b) Also as shown,

x (t ) = (1/2 t ³ + t + 4) i + (t ² - 4t + 2) j

(c) First, find the particle's position at t = 3 :

x (3) = (1/2•3³ + 3 + 4) i + (3² - 4•3 + 2) j = 41/2 i - j

The particle's distanc from the origin is the magnitude of this vector:

|| x (3) || = √((41/2)² + (-1)²) = √1685/2 ≈ 20.52