Department of Physics and Applied Physics 95.141, F2010, Lecture 4 Physics I 95.141 LECTURE 4...

Department of Physics and Applied Physics95.141, F2010, Lecture 4

Physics I95.141

LECTURE 49/15/10


Exam Prep Problem

• Vince Carter’s vertical leap is 43”. – A) (10pts) With what initial vertical velocity does Carter

leave the ground?

– B) (10pts) What is his hang time?

– C) (10pts) Assuming Carter leaps straight up at t=0s and lands at just after t=T, draw the vectors for James’ displacement, velocity, and acceleration at:

• i) The instant he leaves the ground (t=0s)

• ii) t=T/4

• iii) t=T/2

• iv) t=3T/4

• v) t=T


Exam Prep Problem

• Vince Carter’s vertical leap is 43”. – A) (10pts) With what initial vertical velocity does

James leave the ground?


Exam Prep Problem

• Vince Carter’s vertical leap is 43”. – B) (10pts) What is his hang time?


Exam Prep Problem

• Vince Carter’s vertical leap is 43”. – C) (10pts) Assuming Carter leaps straight up at t=0s

and lands at just after t=T, draw the vectors for James’ displacement, velocity, and acceleration at:0t 4

Tt 2Tt 4

3Tt Tt

avy

avy

avy

avy

avy


Outline

• Lecture 3 Review• Vector Kinematics• Relative Motion

• What do we know?– Units/Dimensions/Measurement/SigFigs– Kinematic equations– Freely falling objects– Vectors


Lecture 3 Review

• Freely falling body problems– Batman’s bat-hook

• Scalars and Vectors• Graphical description of vectors and vector

addition.• Vector components• Mathematical description of vector addition

(addition of components)• Unit Vectors


Vector Kinematics

• We can now do kinematics in more than one dimension– This is helpful, because we live in a 3D world!

• We previously described displacement as Δx, but this was for 1D, where motion could only be positive or negative.

• In more than 1 dimension, displacement is a vector

v

r


Vector Kinematics12 xxx

12 rrr

Now, instead of describing displacement in terms of either vertical or horizontal position, we can talk about a displacement vector!

x

y


Vector Kinematics

• In unit vectors, we can write the displacement vector as:

• We can now rewrite our expression for average velocity:


Vector Kinematics

• Average velocity only tells part of the story• Just like for motion in 1D, we can let Δt get smaller and smaller….• Gives instantaneous velocity vector:

t

r

tv

0

lim


Velocity Vector

• The magnitude of the average velocity vector is NOT equal to the average speed.

• But the magnitude of the instantaneous velocity vector is equal to the instantaneous speed at that time


Instantaneous Velocity (math)

• To find the instantaneous velocity, we can take the derivative of the position vector with respect to time:

ktzjtyitxr ˆ)(ˆ)(ˆ)(


Example

• Say we are given the position of an object to be:

ktjeittr t ˆ)2sin(ˆ2ˆ)14()( 2

• Can we find the velocity as a function of time?


Acceleration Vector

• Average acceleration:

• Instantaneous acceleration


Example Problem

• Imagine we are given the position of an object as a function of time– Find displacement at t=1s and t=3s– Find velocity and acceleration as a function of time– Find velocity and acceleration at t=3s

jtmitttrs

ms

ms

m ˆ)(3)(3ˆ)(2)(4)( 3232


Example Problem


jtmitttrs

ms

ms

m ˆ)(3)(3ˆ)(2)(4)( 3232


Example Problem



Example Problem

• Let’s say we are told that a Force causes an object to accelerate in the -y direction at 5m/s2. The object has an initial velocity in the +x direction of 10m/s, and in the +y direction of 15 m/s, and starts at the point (0,0).

– A) Give the initial velocity vector of the object

– B) Plot x(t) vs. t

– C) Plot y(t) vs. t

– D) Plot the object’s trajectory in the xy plane


Example Problem• Let’s say we are told that a Force causes an object to

accelerate in the -y direction at 5m/s2. The object has an initial velocity in the +x direction of 10m/s, and in the +y direction of 15 m/s, and starts at the point (0,0)..– A) Give the initial velocity vector of the object

20

20

vx

vy


Example Problem• Let’s say we are told that a Force causes an object to accelerate in

the -y direction at 5m/s2. The object has an initial velocity in the +x direction of 10m/s, and in the +y direction of 15 m/s, and starts at the point (0,0).– Before we solve B-D, let’s determine equations of motion

(METHOD I)



the -y direction at 5m/s2. The object has an initial velocity in the +x direction of 10m/s, and in the +y direction of 15 m/s, and starts at the point (0,0).– Before we solve B-D, let’s determine equations of motion

(METHOD II)



the -y direction at 5m/s2. The object has an initial velocity in the +x direction of 10m/s, and in the +y direction of 15 m/s, and starts at the point (0,0).– B) Plot x(t) vs t

10

100

t(s)

x(t)

time x(t)

0 0m

1 10m

2 20m

5 50m

10 100m

ttvtx ox 10)(



the -y direction at 5m/s2. The object has an initial velocity in the +x direction of 10m/s, and in the +y direction of 15 m/s, starts at (0,0).– C) Plot y(t) vs t

time y(t)

0 0m

1 12.5m

2 20m

3 22.5m

4 20m

5 12.5

10 -100

22 5.2152

1)( ttattvty oy


Example ProblemD) Plot object trajectory

• Choose coordinate system

time x(t) y(t)

0 0m 0m

1 10m 12.5m

2 20m 20m

3 30m 22.5m

4 40m 20m

5 50 12.5

10 100 -100

25.215)( ttty

ttx 10)(

2025.5.1)( xxxy


Relative Velocity

• So far we have looked at adding displacement vectors

• May also find situations where we need to add velocity vectors


Relative Velocity

• Two velocities:

5m/s

25m/s


Relative Velocity

• In this case, our hero would presumably prefer not to be decapitated by the bridge

• So we are interested in his velocity relative to the bridge• He is on a train moving at +25 m/s relative to the bridge• His velocity relative to the train is -5m/s• So his velocity relative to the bridge is:


Relative Velocity

• What is Kirk’s velocity when he hits the ground?– Assume he leaps when car is moving 20m/s


Relative Velocity (Example 1)

• Imagine you are on a barge floating down the river with the current

• You walk diagonally across the barge with a velocity

• What is your velocity with respect to the water?

• With respect to the river bank?

iv sm

river

3

jiv sm

sm ˆ2ˆ2bargeon walk


Another River Problem

• A boat’s speed in still water is 1.85m/s. If you want to directly cross a stream with a current 1.2m/s, what upstream angle should you take?


Today We Learned….

• Vector kinematics– Displacement vector – Average velocity vector– Inst. Velocity vector– Average acceleration vector– Inst. Acceleration vector– Vector equations of motion

• Relative Velocity

Date post:	21-Dec-2015
Category:	Documents
View:	215 times
Download:	1 times

Department of Physics and Applied Physics 95.141, F2010, Lecture 4 Physics I 95.141 LECTURE 4...

Documents