Answer to Question #111645 in Vector Calculus for Felix Coleman

Question #111645
The position vector r of a particle at time t is
r = 2t
2
i + (t
2 − 4t)j + (3t − 5)k.
Find the velocity and the acceleration of the particle at time t. Show that when t =
2
5
the velocity
and the acceleration are perpendicular to each other. The velocity and the acceleration are resolved
into components along and perpendicular to the vector i − 3j + 2k. Find the velocity and acceleration
components parallel to this vector when t =
2
5
1
Expert's answer
2020-04-30T19:11:34-0400

Given "\\vec{r} = 2 t ^2 \\hat{i} +(t^2 - 4t)\\hat{j}+(3t-5)\\hat{k}."

So, Velocity "\\vec{v} = \\frac{d\\vec{r}}{dt} = 4t \\hat{i} +(2t-4)\\hat{j}+3\\hat{k}"

and acceleration "\\vec{a} = \\frac{d\\vec{v}}{dt} = 4\\hat{i} + 2\\hat{j}".

Now velocity and acceleration are perpendicular when "\\vec{v}\\cdot\\vec{a} = 0".

So, "(4t)\\times 4 +(2t-4)\\times 2 = 0, so \\hspace{0.1 in} 20t = 8, so \\hspace{0.1 in} t =\\frac{8}{20}=\\frac{2}{5}."


Now, given another vector is "\\vec{b} = \\hat{i}-3\\hat{j}+2\\hat{k}."

Component of velocity along this

"\\vec{v_a} = \\frac{(\\vec{v}.\\vec{b})\\vec{b}}{|\\vec{b}|^2} = \\frac{(4t-6t+12+6)(\\hat{i}-3\\hat{j}+2\\hat{k})}{1^2+3^2+2^2} = (\\frac{-t+9}{7})(\\hat{i}-3\\hat{j}+2\\hat{k})" .

Component of acceleration along given vector

"\\vec{a_a} = \\frac{(\\vec{a}.\\vec{b})\\vec{b}}{|\\vec{b}|^2} = \\frac{(-1)}{7} (\\hat{i}-3\\hat{j}+2\\hat{k})" .

At t = 2/5, "\\vec{v}_a = \\frac{43}{35} (\\hat{i}-3\\hat{j}+2\\hat{k}), and \\hspace{0.1 in} \\vec{a}_a=\\frac{-1}{7} (\\hat{i}-3\\hat{j}+2\\hat{k})".

Now, Component of velocity perpendicular to given vector = "\\vec{v}-\\vec{v}_a".

And component of acceleration perpendicular to given vector = "\\vec{a} - \\vec{a}_a"


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