Determining velocity/acceleration in vector-unit notation

[mcdowellmg]

stupid, stupid negative sign. Thanks anyway!

 

[Mark44]

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)|.

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.

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!

 

[Mark44]

The angle theta between two vectors u and v can be obtained from this formula.
u \cdot v = |u||v|cos(theta)
==> cos(theta) = (u \cdot 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.