Thread: Finding the Velocity Vector and Speed after t seconds

1. Finding the Velocity Vector and Speed after t seconds

This is to do with mechanics, which as far as I know is pre-calc, not pre-algebra.

Question: A particle has an initial velocity of $\displaystyle i+2j$ and is accelerating in the direction $\displaystyle i+j$. If the magnitude of the acceleration is $\displaystyle 5\sqrt{2}$, find the velocity vector and the speed of the particle after 2 seconds.

Answer:
velocity vector: $\displaystyle 11i+12j$
speed of particle: $\displaystyle \sqrt{265} ms^{-1}$

I am literally stumped on how to achieve this answer. If you can tell me what to do to get there, I will try working it out .

2. Originally Posted by Viral
This is to do with mechanics, which as far as I know is pre-calc, not pre-algebra.

Question: A particle has an initial velocity of $\displaystyle i+2j$ and is accelerating in the direction $\displaystyle i+j$. If the magnitude of the acceleration is $\displaystyle 5\sqrt{2}$, find the velocity vector and the speed of the particle after 2 seconds.

Answer:
velocity vector: $\displaystyle 11i+12j$
speed of particle: $\displaystyle \sqrt{265} ms^{-1}$
direction of the acceleration vector is $\displaystyle \arctan(1) = 45^\circ$

since the acceleration vector has magnitude $\displaystyle 5\sqrt{2}$, each of its components has magnitude $\displaystyle 5$.

$\displaystyle a = 5i + 5j$

$\displaystyle v(t) = (1 + 5t)i + (2 + 5t)j$

$\displaystyle v(2) = 11i + 12j$

$\displaystyle |v(2)| = \sqrt{11^2 + 12^2} = \sqrt{265}$

3. Ehh, that seems too condensed. Can you expand on the v(t) and v(2) parts, and is there a way without using arctan? We haven't had to use trigonometry for these questions, although we have used pythagoras ($\displaystyle \sqrt{i^{2}+j^{2}}$).

I worked out the 5i and 5j components somehow, but that's as far as I got.

4. Originally Posted by Viral
Ehh, that seems too condensed. Can you expand on the v(t) and v(2) parts, and is there a way without using arctan? We haven't had to use trigonometry for these questions, although we have used pythagoras ($\displaystyle \sqrt{i^{2}+j^{2}}$).

I worked out the 5i and 5j components somehow, but that's as far as I got.
you should know that $\displaystyle a = 5i+5j$ forms a 45-45-90 triangle, so there is no need for the arctangent function in this case.

the basic velocity kinematics equation for constant acceleration is $\displaystyle v = v_o + at$

component-wise, for your problem, that would be

$\displaystyle v_x = v_{ox} + a_x t = 1 + 5t$

so, velocity as a function of time in the x-direction is $\displaystyle (1+5t)i$

$\displaystyle v_y = v_{oy} + a_y t = 2 + 5t$

and, velocity as a function of time in the y-direction is $\displaystyle (2+5t)j$

the overall velocity vector represented in component form is

$\displaystyle v(t) = (1+5t)i + (2+5t)j$

that's about as easy as I can put it for you without teaching about two-dimensional kinematics in total.

5. Hmm, I just tried working it out again, without the arctan stuff and got the answer right:

1). $\displaystyle v=u+at=11i+12j$
2). $\displaystyle \sqrt{11^{2}+12^{2}}=\sqrt{265}ms^{-1}$

But now I don't see where the direction of acceleration came into play at all.

6. Originally Posted by Viral
Hmm, I just tried working it out again, without the arctan stuff and got the answer right:

1). $\displaystyle v=u+at=11i+12j$
2). $\displaystyle \sqrt{11^{2}+12^{2}}=\sqrt{265}ms^{-1}$

But now I don't see where the direction of acceleration came into play at all.
what if (i + j) was not the direction of the acceleration vector?

how, then, would one determine the value of each component?

7. That's what I don't understand, since I didn't use it at all (well, I obviously did, but the value of 1 doesn't change my final answer).