# Math Help - Vectors

1. ## Vectors

The position of a particle in space is given by the vector equation: $r(t)= (t+1)^2 i + (4/3)sqrt(19)(t+1)^(3/2) j + 19sqrt(6)(t+14) k$

Determine when will the speed of the particle be 57 units?

Since the given equation is a position vector, I must take its derivative to find the velocity function. From that since its asking for the time to reach a speed of 57 units, i set the V(t) = 57 and solve for t? Would this be the correct way to solve this question?

2. Originally Posted by Belowzero78
The position of a particle in space is given by the vector equation: $r(t)= (t+1)^2 i + (4/3)sqrt(19)(t+1)^(3/2) j + 19sqrt(6)(t+14) k$

Determine when will the speed of the particle be 57 units?

Since the given equation is a position vector, I must take its derivative to find the velocity function. From that since its asking for the time to reach a speed of 57 units, i set the V(t) = 57 and solve for t? Would this be the correct way to solve this question?
You solve $|\vec{v}(t)|$ = 57.

3. Yes, Mr. Fantastic is right, you set the *magnitude* of v(t) to 57. So you have:

$v(t)=(2t+2)\ \mathbf{i}+2\sqrt{19}(t+1)^{1/2}\ \mathbf{j}+19\sqrt{6}\ \mathbf{k}$

If you let s=t+1 (just to make the algebra easier), and set the square of the magnitude of v(t) equal to 57 squared (setting the squares equal just to make the algebra easier), you get:

$4s^2 + 76s + 2166 = 3249$

Which you can solve for s, and then t=s-1.

Post again in this thread if you're still having trouble.

4. Originally Posted by hollywood
Yes, Mr. Fantastic is right, you set the *magnitude* of v(t) to 57. So you have:

$v(t)=(2t+2)\ \mathbf{i}+2\sqrt{19}(t+1)^{1/2}\ \mathbf{j}+19\sqrt{6}\ \mathbf{k}$

If you let s=t+1 (just to make the algebra easier), and set the square of the magnitude of v(t) equal to 57 squared (setting the squares equal just to make the algebra easier), you get:

$4s^2 + 76s + 2166 = 3249$

Which you can solve for s, and then t=s-1.

Post again in this thread if you're still having trouble.
Alright, so i rearranged the equal so now it will be $4s^2 +76s -1083 = 0$.

This equation factors to: $(2s - 19)(2s + 57)$

Solving for s we get: $s = \frac{19}{2}, s = \frac{-57}{2}$

I took the positive s which indeed evaluates to $\frac{19}{2} = t + 1$.

Solving for $t$, I indeed get $t = \frac{17}{2}$.

Thanks hollywood!