Velocity Vector at the end of the acceleration formula?

[NiendorfPhysics]

Homework Statement

Don't ruin the problem for me please, I just want to know why there is a velocity vector at the end of this formula. It's really strange since there is velocity in the equation already, so what would the velocity with the vector arrow on top mean?

Problem: A boat travels at constant velocity ⃗vo when the motor is on. At t = 0 the motor is turned off and the acceleration of the boat due to the resistive force of the water is ⃗a =−kv^(−1/2⃗)v (with k > 0 a constant). a) Find ⃗a(t), ⃗v(t) and ⃗r(t). b) Find the time and the distance to stop.

PDF source (problem 4): http://web.physics.ucsb.edu/~physCS31/fall2014/hw3.pdf

Homework Equations

a (with vector arrow) = −kv^(−1/2⃗)v

The Attempt at a Solution

I thought at first that it might be like the r-hat directional vector at the end of some equations, but now I'm not so sure

 

[gneill]

Is that:
$\left(-k v^{-1/2}\right) \vec{v}~~~~~~~~~~$ or $~~~~~~~~~~~\left(\frac{-k v^{-1}}{2}\right) \vec{v}$ ?

 

[NiendorfPhysics]

Is that:
$\left(-k v^{-1/2}\right) \vec{v}~~~~~~~~~~$ or $~~~~~~~~~~~\left(\frac{-k v^{-1}}{2}\right) \vec{v}$ ?

The first one. Sorry, I have no clue how to format equations on a computer

 

[gneill]

The first one. Sorry, I have no clue how to format equations on a computer

Okay, well I guess we'll need to see some context for the equation: how it came about. If it was the second version then your idea of it being essentially a directional unit vector would make sense since $v$ would be the magnitude of $\vec{v}$ and since $v^{-1}$ is just $1/v$ you'd have a vector divided by its magnitude, i.e., a unit vector in the direction of $\vec{v}$.

 

[NiendorfPhysics]

Okay, well I guess we'll need to see some context for the equation: how it came about. If it was the second version then your idea of it being essentially a directional unit vector would make sense since $v$ would be the magnitude of $\vec{v}$ and since $v^{-1}$ is just $1/v$ you'd have a vector divided by its magnitude, i.e., a unit vector in the direction of $\vec{v}$.

The equation was just given to me in that form within the question. Also it says k is a constant >0. Umm as for other context I'm not really sure what else I could say. It asks me to solve for a(t) v(t) and r(t) (all with the vector arrow) and to find the time and distance to stop, but those two questions should be relatively easy once I figure out how to interpret the equation

 

[gneill]

I see. I don't know what more I can contribute here, since at face value the units don't work out to be acceleration if the constant k is a simple numerical value. Can you post the original question that this came from?

 

[NiendorfPhysics]

I see. I don't know what more I can contribute here, since at face value the units don't work out to be acceleration if the constant k is a simple numerical value. Can you post the original question that this came from?

4. A boat travels at constant velocity ⃗vo when the motor is on. At t = 0 the motor is turned off and the acceleration of the boat due to the resistive force of the water is ⃗a = −kv^(−1/2⃗)v (with k > 0 a constant). a) Find ⃗a(t), ⃗v(t) and ⃗r(t). b) Find the time and the distance to stop.

Or if you would like the pdf that it came from (problem 4) http://web.physics.ucsb.edu/~physCS31/fall2014/hw3.pdf

 

[gneill]

I think you'll have to presume that the constant k will have suitable units associated with it in order for the equation to balance. If you multiply through by v/v you can associate the v in the denominator with the vector $\vec{v}$ to form the unit vector. According to that the acceleration vector has the same direction as the velocity vector, and its magnitude be proportional the the square root of v. Looks like a differential equation...