Two Dimensional Motion

ScalarvsVector

Scalar - quantity specified by only a number with appropriate units

Vector - quantity that has magnitude and direction.

Vector Addition

Triangle Method

Tail of one vector is placed at the head of the other.

The resultant is the vector drawn from the tail of the first to the head of the last vector.

Have you ever tried to ride a bike in strong wind?

A cyclist is riding due east at 16.2 km/hr. Consistent wind traveling due west is recorded at 16.3 km/hr. What is the resultant velocity of the cyclist?

Cyclist at 16.2 km/hr

Wind at 16.3 km/hr

Cyclist velocity is .1 km/hr due WestOr

-.1 km/hr due East

Ratios to show expressions

A luge is on a track moving 240 km/hr. When it turns a corner there is a headwind (exact opposite direction) of 60km/hr. If Vluge is used to represent the original vector velocity. Write an expression for the resulting velocity in the headwind in terms of Vluge

We know that we can add the two vectors to find the Velocity of the plane. 240 + -60 = 180m/s

Then we can make a ratio of the resulting velocity to the original velocity to represent an expression for the vector.

= Vluge

The luge is traveling ¾ of its original velocity.

Resultant Vectors

Magnitude

Direction

How do we find can be used to find the magnitude and direction of the resultant vector.

Pythagorean theorem and the inverse tangent function

d (displacement) =

Tangent of angle

= 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆 𝒍𝒆𝒈

𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒍𝒆𝒈

-1 𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆

𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕

d =

d = d=

d = d = 15.0 blocks

You can easily see that direction is south east when looking at

dd

37°

Displacement in Two Dimensions

dy = vertical displacement dx = horizontal displacement ax =horizontal acceleration ay = vertical acceleration

= horizontal velocity ( )

= vertical velocity ( )

i - initial f - final can also be written just as t

Vectorsto Components (vertical motion and horizontal motion)

Sine and Cosine functions can also be used to find components of vectors.

y

Launched at an angle

Be Careful!

Physics Club is launching watermelons to hit a target 4.5m high. The release of the sling shot is 1.0m above ground level and are set up 15.0m from the target. What launch angle should students try and achieve? What initial velocity is needed to hit target in 3.0 seconds?

15.0m 1.0m

3.5m

=

Ideal launch angle =13°

= =

= 1.7m/s

An older sister pushes her unsuspecting sister off a pier at Lake Tahoe. What horizontal velocity did the younger sister travel with if the pier was 25ft high and she landed 1.8 meters away?

1.8 m

25ft

? m/sFirst order of business?? Make

sure all units match

25ft x .

= 7.6 m

vy = 0.0m/sdy = 7.6m

g = 9.81 m/s2

dx = 1.8 m

vx = ?

If gravity is noted specifically as a positive

value, then vertical displacement is also

positive!

Δ𝑡 =2𝑑

𝑎

=

= 15.2/9.81=1.2447

= /

= 1.8 / 1.2447

= 1.446=1.4 m/s

Hint!!! - Many of the questions Quiz 2 use these two equations. You will just be

solving for different portions.

2

Vy = 0 so we can remove terms

2

Solve for

Putting it all together!

A cliff Jumper runs horizontally off a cliff with velocity of 2.24 m/s. The cliff is 18.0m high. The boat is 9.0 m from the base of the cliff. The boat wants to be only .5 meter from the jumper when he hits the water. What will the average velocity of the boat need to be?

2.24 m/s

18.0m

.50 m

Vy = 0.0 m/s-18.0m/s

a = g = -9.81 m/s2

?Vx = 2.24 m/s

?

=

=

= 1.91565

dx = jumper landing

= 2.24 (1.91565)= 4.29m

Boat to landing pt= 9.0m – (4.29+.50)

= 4.21m

vavg boat =

=

9.0m

vavg = 2.2 m/s

Relative Velocity

Video

https://www.khanacademy.org/science/in-in-class11th-physics/in-in-class11th-physics-motion-in-a-straight-line/in-in-relative/v/calculating-relative-velocity-class-11-india-physics-khan-academy

or google search Khan Academy calculating relative velocity