4.1.1 Write the vector equation of the position of a particle that moves in a ln

Question

4.1.1 Write the vector equation of the position of a particle that moves in a lne, has constant velocity, is at (2,5) at time zero and is at (1, 4) at time one. What is the slope of this line? 4.1.2 Write the vector equation of the position of a particle that moves in a line, has constant velocity, is at (0,-3) at time zero and is at (2,3) at time two. What is the slope of this line? 4.1.3 A boat is initially one mile east and three miles north of the harbor. The coordinates of the port are taken to be (0,0) and north is taken to be in the direction of the positive y axis. The boat is moving at a constant velocity of three miles per hour north and one mile per hour east. Find the equation for the boat's position as a function of time and draw the path of motion of the boat on the coordinate plane. What is the boat's speed? 4.1.4 A line L intersects the points (3, 3) and (5,7). What is the closest point on L to the point (1,2)? 4.1.5 The line L has slope equal to 3. slope of L1? he line Li is perpendicular to L. What is the 4.1.6 The line L intersects (2,7) and (5, 1). Find an equation for the line perpendicular to L and intersecting (2,1). Sketch the graph of the two lines. Where do they intersect? 4.1.7|| Boat A is initially at position (2,5) and moves at a constant velocity of 4.1) Boat B s at position (8,3) and moves at a constant velocity of 2.10) (a) Do the paths of the boats ever cross? If so where? (b) Do the boats collide? (c) If the boats do not collide, what is the minimum distance between the boats and at what time are the boats this minimum distance apart?

Explanation / Answer

(2,5) and (1,4)

So, p = (2,5) + t<(1,4)- (2,5)>

p = (2,5) + t<-1,-1>

p = <2 - t , 5 - t>

So, the vector equation for position is :
x = 2 - t
y = 5 - t

When we eliminate parameter t, we get :
y - x = 3
y = x + 3

So, comparing with y = mx + c, we get
slope, m = 1

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(0,-3) at t = 0
(2,3) at t = 2

x = 0 when t = 0
x = 2 when t = 2
So, x = t

y = -3 when t = 0
y = 3 when t = 2
So, using slope = (3 - (-3))/(2 - 0) --> 3
y = 3t + B
And using (0,-3), we get B = -3
So, y = 3t - 3

And thus position is
x = t
y = 3t - 3

Eliminating t :
y = 3x - 3
So, slope =3

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4)
(3,3) and (5,7) line :
Slope = 2
y = 2x + b
And using (3,3), we get b = -3
So, y = 2x - 3

This has slope = 2
So, perpendicular line to this line will have slope = -1/2

y = -0.5x + C
And using (1,2), we get C = 2.5

So, y = -0.5x + 2.5

Now, solving
y = 2x - 3
y = -0.5x + 2.5

Equating :
2x - 3 = -0.5x + 2.5
2.5x = 5.5
x = 2.2
And solving, we get y = 1.4

So, pt of intersection is (2.2 , 1.4)

Now distance b/w (1,2) and (2.2,1.4) is :
sqrt(1.2^2 + 0.6^2)
sqrt(1.44 + 0.36)
sqrt(1.80)
= 1.34164 ---> ANS

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5)
Slope = 3
We know that for perp lines, product of slopes = -1
m1 * m2 = -1
3 * m2 = -1

So, m2 = -1/3

So slope of perp line is -1/3

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