Simple Nature - Light and Matter

aa/Visualizing the accelera-
tion vector.

Incorrect solution #4:

(same notation as above)

Ax= (5.0 km) cos 45◦= 3.5 km

Bx=−(12.0 km) cos 55◦=−6.9 km

Ay= (5.0 km) sin 45◦= 3.5 km

By=−(12.0 km) sin 55◦=−9.8 km

Cx=Ax+Bx

=−3.4 km

Cy=Ay+By

=−6.3 km

|C|=

√

Cx^2 +C^2 y

= 7.2 km

direction = tan−^1 (−6.3/−3.4)

= 62◦north of east

Incorrect solution #5:

(same notation as above)

Ax= (5.0 km) cos 45◦= 3.5 km

Bx=−(12.0 km) sin 55◦=−9.8 km

Ay= (5.0 km) sin 45◦= 3.5 km

By=−(12.0 km) cos 55◦=−6.9 km

Cx=Ax+Bx

=−6.3 km

Cy=Ay+By

=−3.4 km

|C|=

√

Cx^2 +C^2 y

= 7.2 km

direction = tan−^1 (−3.4/−6.3)

= 28◦north of east

3.4.4 Calculus with vectors

Differentiation

In one dimension, we define the velocity as the derivative of the

position with respect to time, and we can think of the derivative as

what we get when we calculate ∆x/∆tfor very short time intervals.

The quantity ∆x=xf−xiis calculated by subtraction. In three

dimensions,xbecomesr, and the ∆rvector is calculated byvector

subtraction, ∆r=rf−ri. Vector subtraction is defined component

by component, so when we take the derivative of a vector, this means

we end up taking the derivative component by component,

vx=

dx

dt

, vy=

dy

dt

, vz=

dz

dt

212 Chapter 3 Conservation of Momentum