Parametric Vectors

• Apr 12th 2011, 08:35 AM
JewelsofHearts
Parametric Vectors
Let p(t) give the distance in meters from (0, 1) at time t.

$\displaystyle p(t)= <2t,cos(t)>$

(a) Find the acceleration vector and,
(b) use it to calculate the acceleration at t = 5p/ 6 .

Explain what the acceleration of the particle means in relation to its position at t = 2.

I have differentiated the vectors to get:

v(t)=<2,-sin(t)>
for velocity

and

a(t)<0,-cos(t) >
for acceleration.

-cos(5pi/6)=0.866. Should this be the final answer for acceleration or should the answer be written in vector format? When the problem asks for acceleration, I believe it asking for the net acceleration, so that is why I used the pythagoream theorem to get 0.866.

For the part of the problem where it asks to explain what the relationship between the accel. and pos. at t=2, I don't if I should compare the the accel. vector with the veloc. vector or compare the net accel. and net veloc.

Help, please. I know it's easy. I just don't know what they're asking for.
• Apr 12th 2011, 09:25 AM
HallsofIvy
Quote:

Originally Posted by JewelsofHearts
Let p(t) give the distance in meters from (0, 1) at time t.

$\displaystyle p(t)= <2t,cos(t)>$

(a) Find the acceleration vector and,
(b) use it to calculate the acceleration at t = 5p/ 6 .

Explain what the acceleration of the particle means in relation to its position at t = 2.

I have differentiated the vectors to get:

v(t)=<2,-sin(t)>
for velocity

and

a(t)<0,-cos(t) >
for acceleration.

-cos(5pi/6)=0.866. Should this be the final answer for acceleration or should the answer be written in vector format? When the problem asks for acceleration, I believe it asking for the net acceleration, so that is why I used the pythagoream theorem to get 0.866.

Only your teacher can say for sure what is wanted- I would recommend you give both. By the way, I would use the exact value of $\displaystyle \sqrt{3}/2$ rather than the approximate value of 0.866.

Quote:

For the part of the problem where it asks to explain what the relationship between the accel. and pos. at t=2, I don't if I should compare the the accel. vector with the veloc. vector or compare the net accel. and net veloc.
You mean position, not velocity. As you say, the acceleration vector, at any time, is <0, -cos(t), and you are given that the position vector is <2t, cos(t)>. At t= 2, those are <0, -cos(2)> and <4, cos(2)>. I can see no good reason to compare $\displaystyle \sqrt{cos^2(2)}= |cos(2)|$ and $\displaystyle \sqrt{4+ cos^2(2)}$.

Quote:

Help, please. I know it's easy. I just don't know what they're asking for.