Kinematics in 2 or 3DChapter 3

VectorsMany of the quantities we have been discussing are vectors. If a quantity is a vector or not is very important in understanding what the quantity means and how to use it.

A vector is something that has both a magnitude and a direction.

Something which only has magnitude but no direction is called a scalar.

Vectors

What quantities that we’ve discussed so far are vectors?

Which are scalars?

distance velocity acceleration

time

Adding Vectors

The main feature of vectors that makes them more difficult is that you can’t simply add them as numbers.

You all know how to add 5+5, but how about 5 km north + 5 km west?

Multiplication is even worse, what is 5 km north times 10 km west? For now we will only worry about addition, but we will eventually deal with multiplication too.

Graphical MethodsOne way to visualize adding vectors is to draw them as arrows.

To add two vectors you place the tip of one on the tail of the other.

+ =

- =

+ =

notice it doesn’t matter what order you add the vectors

You aren’t going to add vectors this way, but this should help you visualize what you expect your answer to be and avoid mistakes

Vectors by ComponentsThe real way to deal with vectors is to deal with their components.

Once you realize that only the components matter, everything with vectors just becomes dealing with the components.

~D

Dx

Dy

✓

Dx = D cos ✓

Dy = D sin ✓

then you can add the x and y components separately

~D = ~A+ ~B

Dx = Ax +Bx

Dy = Ay +By

Components to VectorsIf you have the components and want the vector back you can use the following two equations:

D =q

D2x +D2

y

Dy

Dx

~DDy

Pythagorean Theorem

Definition of Tangent✓

✓ = tan�1�Dy

Dx

�

Example 3-2A rural mail carrier leaves the post office and drives 22.0 km in a northerly direction. She then drives in a direction 60.0˚ south of east for 47.0 km. What is her displacement from the post office?

x

y

~D1 ~D2

60˚

~D

✓

break into components

D2x = (47.0km)(cos(60)) = (47.0km)(0.50) = 23.5km

D2y = (�47.0km)(sin(60)) = (�47.0km)(0.87) = �40.7km

D1x = 0

D1y = 22.0km

Dx = D1x +D2x = 0 + 23.5km = 23.5km

Dy = D1y +D2y = 22.0km� 40.7km = �18.7km

|D| =q

D2x +D2

y =p(23.5km)2 + (�18.7km)2 = 30.0km

✓ = tan�1�Dy

Dx

�

= tan�1��18.7km

23.5km

�

= �38.5�

Example 3-3An airplane trip involved 3 legs, with two stopovers. The first leg is due east for 620km; the second leg is southeast (45˚) for 440km; and the third leg is 53˚ south of west for 550km. What is the plane’s total displacement?

x

y~D1

~D2

~D3

~DR

45˚

53˚

to solve just break into components

D1x = 620km D1y = 0km

D2x = (440km) cos(45) = 311kmD2y = (440km) sin(45) = �311km

D3y = (�550km) sin(53) = �439km

D3x = (�550km) cos(53) = �331km

Dx = D1x +D2x +D3x = 620km+ 311km� 331km = 600km

Dy = D1y +D2y +D3y = 0km� 311km� 439km = �750km

DR =qD2

x +D2y =

p(600km)2 + (�750km)2 = 960km

tan ✓ =Dy

Dx=

�750km

600km= �1.25

=> ✓ = �51�

θ

Unit VectorsSometimes it is useful to indicate the direction of a vector with unit vectors.

A unit vector is just a vector that has a length of 1, so basically it has direction but not magnitude.

You can then right a vector as:

~V = Vxi+ Vy j + Vz k ~V = Vxx+ Vy y + Vz zor

Engineering often uses i, j, k to stand for x, y, z

Vector KinematicsThe key point of all this is that the kinematics we have learned for 1D, work just as well in 3D.

We can write the equations as vectors:

However, it practice we will always solve things by using the vector components.

~v =d~r

dt~a =

d~v

dtNote we will now use r for

a displacement in 3D

~v = vxx+ vy y + vz z =dx

dtx+

dy

dty +

dz

dtz

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~a = axx+ ay y + az z =dvxdt

x+dvydt

y +dvzdt

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OrthogonalThe key thing is that the x,y and z directions are orthogonal.

What that means is that they have nothing to do with one another. You only have to consider each component separately.

ax =dvxdt

ay =dvydt

az =dvzdt

vz =dz

dt

vy =dy

dt

vx =dx

dtThis is just what we have been doing.

In 2D or 3D you just do the same thing, but for the other components.

Projectile MotionProjectile motion is the study of objects that have constant acceleration in one direction and an initial velocity.

This corresponds to lots of things in the real world; a cannon ball, a tennis ball, jumping out of a plane, golf, a water fountain, the high jump, a coyote running off of a cliff.

The trajectory of projectile motion is a parabola.

Example 3-6A movie stunt driver on a motorcycle speed horizontally off a 50.0m high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0m from the base of the cliff where the cameras are? Ignore air resistance.

x

y

50.0m

90.0m

knownx0 = 0y0 = 0

x = 90.0my = -50.0m

ax = 0ay = -9.81m/s2

unknown

vyo = 0

vx0 = ?

physicsx:x = x0 + vx0t+

1

2axt

2 = vx0t => vx0 =x

t

y:y = y0 + vy0t+

1

2ayt

2 =1

2ayt

2 => t =

s2y

ay=> t =

s2(�50.0m)

�9.81m/s2= 3.19s

vx0 =90.0m

3.19s= 28.2m/s

Example 3-7A football is kicked at an angle θ0=37.0˚ with a velocity of 20.0 m/s. Calculate a) the maximum height, b) the time of travel before the football hits the ground, c) how far away it hits the ground, d) the velocity vector at the maximum height, and e) the acceleration vector at maximum height. Assume the ball leaves the foot at ground level, and ignore air resistance and rotation of the ball.

x

yknown

x0=0 y0=0vx0=(20.0m/s)cos(37.0) = 16.0 m/s

vy0=(20.0m/s)sin(37.0) = 12.0 m/s ax=0 ay=-9.80m/s2

physics? vy=0vy = vy0 + ayt => t =

�vy0ay

=�12.0m/s

�9.80m/s2= 1.22s

unknowny=?

y = y0 + vy0t+1

2ayt

2 = 0 + 12.0m/s(1.22s) +1

2(�9.80m/s2)(1.22s)2 = 7.35m

Example 3-7A football is kicked at an angle θ0=37.0˚ with a velocity of 20.0 m/s. Calculate a) the maximum height, b) the time of travel before the football hits the ground, c) how far away it hits the ground, d) the velocity vector at the maximum height, and e) the acceleration vector at maximum height. Assume the ball leaves the foot at ground level, and ignore air resistance and rotation of the ball.

b) y = 0 t = ? y = y0 + vy0t+1

2ayt

2 => vy0t+1

2ayt

2 = 0

t(vy0 +1

2ayt) = 0 or t = 0s

c) t = 2.45s x = ? x = x0 + vx0t+1

2axt

2 = (16.0m/s)(2.45s) = 39.2m

d) vy=0 vx=16.0m/s ~v = (16.0m/s) i

e) ax=0 ay=-9.80m/s2 ~a = (�9.80m/s2) j

=2(12.0m/s)

9.80m/s2= 2.45s=> t =

2vy0�ay

The Blues Brothers

This is the 95th St. Bridge in Chicago which we can look up and find has a length of 104.5m

Let’s say the gap between the bridge is 32 m. Then we have

Then the angle is arccos(36.25/51.25) = 45˚. We then have

32m

104.5m36.25m

51.25m

x = 32m

y = 0

vx0 = v0 cos (45)

vy0 = v0 sin (45)

ax = 0

ay = g = �9.81m/s2

y = vy0t+1

2ayt

2 = 0 => vy0 = �1

2ayt => v0 = �1

2

aysin(45)

t

x = vx0t => t =x

vx0=

x

v0 cos(45)=> v0 = �1

2

aysin(45)

x

v0 cos(45)

=> v20 = �1

2

�9.81m/s2

sin(45)

32m

cos(45)= 314m2/s2 => v0 = 17.7m/s

or 40mph

Relative Velocity

The relative velocity between an object and a moving coordinate axis can easily be found by addition or subtraction if they are in the same line.

But if they are not in the same direction, then they must be added as vectors.

Basically we have focussed on how displacement is a vector. But remember velocity and acceleration are too.

Example 3-14A boat’s speed in still water is vBW = 1.85 m/s. If the boat is to travel directly across a river whose current has a speed vWS=1.20 m/s, at what upstream angle must the boat head?

VBS

VWS

VBWθy:

=1.20m/s

1.85m/s= 0.6486sin ✓ =

vWS

vBW

vBW sin ✓ = vWS

✓ = 40.4�

x: what is the speed of the boat with respect to the shore?

vBS = vBW cos ✓ = (1.85m/s)(cos 40.4) = 1.41m/s

Note θ is the angle with the across component so that will be x and the other direction y.

2 Fast 2 Furious

Is this possible?

Homework

Chapter 2: 30,55,64

Chapter 4: 26,39,50,61