Determining velocity/acceleration in vector-unit notation

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In summary, the conversation discusses an electron's position and velocity in unit-vector notation, with a focus on finding its velocity at a specific time and the magnitude and angle of its velocity vector. The solution involves finding the derivative of the position function, calculating the magnitude and unit vector of the velocity, and using the dot product to find the angle.
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mcdowellmg
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stupid, stupid negative sign. Thanks anyway!
 
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mcdowellmg said:

Homework Statement



An electron's position is given by r = 3.00t i - 3.00t2 j + 3.00 k, with t in seconds and r in meters.
(a) In unit-vector notation, what is the electron's velocity v(t)?
v(t) = m/s

(b) What is v in unit-vector notation at t = 4.00 s?
v(4.00) = m/s


(c) What is the magnitude of v at t = 4.00 s?
m/s

(d) What angle does v make with the positive direction of the x-axis at t = 4.00 s?
° (from the +x axis)

Homework Equations



The given r is a function of time t. Velocity is the time derivative of r. The derivative is taken component by component, and the unit-vector symbols must be retained. The magnitude of a vector is calculated with the Pythagorean Theorem, and the angle is calculated with an inverse tangent.


The Attempt at a Solution



I derived the function to 3.00i - 6.00tj. That should be the answer to (a), but it isn't.
The problem asks for a unit vector for the velocity, not just the velocity vector. To make your vector a unit vector, find the magnitude |v(t)|, then multiply by 1/|v(t)|.
mcdowellmg said:
For (b), I have 3.00i + 24.00j (just multiplied 6 by 4 for t).
Your unit vector will be a function of t. Evaluate it at t = 4 seconds.
mcdowellmg said:
I actually got (c) right by doing v = square root of vx^2 + vy^2 (which was the square root of 3^2 + 24^2), getting 24.1868.

For (d), I tried the function theta = inverse tangent(vy/vx), but keep getting an angle of 1.44644 (not right), when I do inverse tangent of 8 (from 24/3).

Thanks for any help!
 
  • #3
The angle theta between two vectors u and v can be obtained from this formula.
u [itex]\cdot[/itex] v = |u||v|cos(theta)
==> cos(theta) = (u [itex]\cdot[/itex] v)/(|u||v|)

Since u and v are (or are supposed to be) unit vectors, the cosine of the angle between them is just their dot product.
 

1. What is velocity in vector-unit notation?

Velocity in vector-unit notation is a way of representing the speed and direction of an object's motion. It is typically written in the form of a vector, with an arrow indicating the direction and a magnitude representing the speed.

2. How is velocity calculated in vector-unit notation?

In vector-unit notation, velocity is calculated by dividing the displacement vector (change in position) by the time interval. This can be represented mathematically as v = Δx/Δt, where v is velocity, Δx is displacement, and Δt is time.

3. What is the difference between velocity and acceleration in vector-unit notation?

Velocity and acceleration are both vector quantities, but they represent different aspects of an object's motion. Velocity is the rate of change of an object's position, while acceleration is the rate of change of an object's velocity. In other words, velocity describes how an object is moving, while acceleration describes how an object's movement is changing.

4. Can acceleration be negative in vector-unit notation?

Yes, acceleration can be negative in vector-unit notation. A negative acceleration indicates that an object is slowing down, while a positive acceleration indicates that an object is speeding up. The sign of acceleration depends on the direction of the object's motion and the direction of its acceleration.

5. How is acceleration represented in vector-unit notation?

Acceleration is represented as a vector in the same way as velocity, with an arrow indicating the direction and a magnitude representing the rate of change. The units for acceleration are typically meters per second squared (m/s^2) in the SI system.

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