VECTOR FUNCTIONS
VECTOR FUNCTIONS
Motion in Space: Velocity and Acceleration
In this section, we will learn about:
The motion of an object using tangent and normal vectors.
MOTION IN SPACE: VELOCITY AND ACCELERATION
Here, we show how the ideas of tangent and normal vectors and curvature can be
used in physics to study:
The motion of an object, including its velocity and acceleration, along a space curve.
VELOCITY AND ACCELERATION
In particular, we follow in the footsteps of
Newton by using these methods to derive
Kepler’s First Law of planetary motion.
VELOCITY
Suppose a particle moves through space so that its position vector at
time t is r(t).
VELOCITY
Vector 1
Notice from the figure
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So, C=i–j+ k
VELOCITY & ACCELERATION
Example 3
It follows: v(t) = 2t2 i + 3t2 j + t k + i – j + k
= (2t2 + 1) i + (3t2 – 1) j + (t + 1) k
VELOCITY & ACCELERATION
Example 3
Since v(t) = r’(t), we have: r(t) = ∫ v(t) dt
= ∫ [(2t2 + 1) i + (3t2 – 1) j + (t + 1) k] dt = (⅔t3 + t) i + (t3 – t) j + (½t2 + t) k + D
VELOCITY & ACCELERATION
Example 3
Putting t = 0, we find that D = r(0) = i.
So, the position at time t is given by: r(t) = (⅔t3 + t + 1) i + (t3 – t) j + (½t2 + t) k
VELOCITY & ACCELERATION
The expression for r(t) that we obtained
in Example 3 was used to plot the path of the particle here for 0 ≤ t ≤ 3.
VELOCITY & ACCELERATION
In general, vector integrals allow us to recover:
Velocity, when acceleration is known
v(t )
v(t0 )
t t0
a(u ) du
Position, when velocity is known
r (t ) r (t0 )
t t0
v(u ) du
VELOCITY & ACCELERATION
If the force that acts on a particle is known, then the acceleration can be found from
Newton’s Second Law of Motion.
VELOCITY & ACCELERATION
The vector version of this law states that if,
at any time t, a force F(t) acts on an object of mass m producing an acceleration a(t),
then
F(t) = ma(t)
VELOCITY & ACCELERATION
Example 4
An object with mass m that moves in
a circular path with
I found the secret formula, it was (w+L)-2 but w/l had to be reduced so it
A(-2, 2) becomes A'(5, 1) , B(-2, 4) becomes B'(5, 3) , C(2, 4) becomes C'(9, 3) , D(2, 2) becomes D'(9, 1)
3 x 2 + -2 x 4 + 3 x 8 + -2 x 1= 6 -8 +24 -2= 20
Ni 2+ (aq) + 2 NO3- (aq) + 2 Na+ (aq) + CO3 2- (aq) NiCO3 (s) + 2 Na+ (aq) + 2 NO3- (aq)
3.1 Using the program shown in Figure 3.30 explain what the output will be at Line A.
= (2t + 2h – t 2 – 2th – h 2 – 2t + t 2 ) = //2t – 2t = 0; t2– t2 =0
Indexi = 0.053 - 0.095 Fi + 0.020 PHDi + 0.007 FPHDi + 0.0015 TENi
87.045 = [d] x 101 + [e] x 100 + [f] x 10-1 + [g] x 10[h] + 5 x 10[i]
2) Calculate following values recursively. ae + bg, af + bh, ce + dg and cf + dh.
When I read through the list of variables, I settled on the very last one, N=11, and divided 36 by it to get 3r3. I, then, scanned back over the list of values to find the corresponding variable for 3 to add to 33 and balance the equation.
help define the number, size, and phasing of the increments. Involving the IPT can reduce
I΄(t) + (a+bt) I(t) = –[ α + β I(t) + γt]with I(t1)=0,0 ≤ t ≤ t1
2(X4 + X5 + X6 + X7 + X10 + X11 + X16 + X17 + X18 + X27 + X28) – (X1 + X2 +X3
Sum: (2x2 1) + (x3 x2 + 2x 1) = x3 + x2 + 2x -2
–x3 – x5 + x13 + x14 + x22 + x23 + x24 = 0