Velocity Vector at the end of the acceleration formula?

• NiendorfPhysics
In summary, the problem presents a boat traveling at constant velocity when the motor is on and experiencing an acceleration due to the resistive force of water when the motor is turned off. The equation for this acceleration is given as ⃗a = −kv^(−1/2⃗)v (with k > 0 a constant). The task is to find ⃗a(t), ⃗v(t), and ⃗r(t) and then determine the time and distance it takes for the boat to stop. The equation appears to be a differential equation and suggests that the acceleration vector has the same direction as the velocity vector and is proportional to the square root of the velocity. The constant k is assumed to have suitable
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

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

gneill said:
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

NiendorfPhysics said:
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}##.

gneill said:
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

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?

gneill said:
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

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...

1. What is the purpose of the velocity vector at the end of the acceleration formula?

The velocity vector at the end of the acceleration formula is used to calculate an object's speed and direction of motion after experiencing a change in acceleration. It is a vector quantity that takes into account both the magnitude and direction of the object's velocity.

2. How is the velocity vector calculated in the acceleration formula?

The velocity vector at the end of the acceleration formula is calculated by multiplying the acceleration vector by the time interval and adding it to the initial velocity vector. This gives the final velocity vector at the end of the acceleration.

3. What is the difference between velocity and acceleration?

Velocity refers to the rate of change of an object's position with respect to time. It includes both the speed and direction of motion. Acceleration, on the other hand, refers to the rate of change of an object's velocity with respect to time. It is a vector quantity that takes into account the change in speed and direction of motion.

4. How is the velocity vector affected by changes in acceleration?

The velocity vector is directly affected by changes in acceleration. An increase in acceleration will result in a larger change in the object's velocity, while a decrease in acceleration will result in a smaller change in velocity. The direction of the velocity vector will also change according to the direction of the acceleration.

5. Can the velocity vector be negative?

Yes, the velocity vector can be negative. This indicates that the object is moving in the opposite direction of the initial velocity vector. The magnitude of the velocity vector is still important in determining the speed of the object, regardless of its direction.

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