Position is specified by x (or x and y in two dimensions)

Displacement is change in position x = xf xi

Distance travelled is length of path

Average velocity is displacement divided by time interval vave = x/t a vector

Average speed is distance travelled divided by time interval a scalar

Instantaneous velocity is limit of velocity as time interval t approaches 0

V = lim x/t

If velocity is constant xf = xi + vt

Average Acceleration is change in velocity divided by time interval aave = v/t a vector

Instantaneous acceleration is limit as time interval t approaches 0

a = lim v/t

If acceleration is constant vf = vi + at

xf = xi + vit + ½ a t2

vf2 = vi

2 + 2 a (xf xi)

The UW is hosting a Public Lecture series on sustainable energy in the Spring quarter.

The lectures will address:

*Solar power (Thursday April 1 at 6:30 PM Kane Hall )*National- and Global-scale planning for sustainable energy*The next generation of nuclear power plants in the U.S.*Genetically-engineered organisms for biofuel production*The environmental impact of sustainable energy projects

Full details are at http://courses.washington.edu/efuture

Details for the 1-2CR seminar course, which will cover a wider range of topics, are at:http://courses.washington.edu/efuture/academic.html

Contact Jerry Seidler at efuture@uw.edu with any questions.

Problem 2-83 deceleration a = -7 m/s2Parameters: reaction time tR = 0.5stopping distance d = 4 mInitial velocity v

Distance travelled during reaction time xR = v tR

Distance travelled during breaking time = xB d = xR + xB

Relation between xB and initial velocity v2 = 2 |a| xB

xB= ½ v2 /ad = xR + xB = vtR + ½ v2/a

v2 +2a tR v – 2a d = 0

A projectile is fired at If air resistance is neglected, the line in the graph that best represents the horizontal displacement of the projectile as a function of travel time is

an angle of 45º above the horizontal

A) 1 B) 2 C) 3 D) 4 E) None of these is correct

target

yT = y1T 1/2 g t2 = xT tan 1/2 g t2

projectile

yP = vyP t 1/2 g t2 = (vP sin t 1/2 g t2

xP = vxP t = (vP cos ) t

t = xP/(vP cos )

yP = (vP sin ) xP/(vP cos ) 1/2 g t2

= xP tan 1/2 g t2

Conclusion: when xP = xT yP = yT

xT

y1T

A particle is moving. At t=0 the velocity vector is v1

A little later the velocity vector is v2The magnitudes are the same: |v1| = |v2|.

v1

v2

The change in velocity is approximately:

A B

C D

E. zero

A particle is moving clockwise around a circle with constant speed.The acceleration vector points inward.

Suppose the particle were movingCounter- Clockwise.The acceleration vector points:

A. InwardB. Outward

The circle has radius R

L is length of a chord; s is arc length

R

Approximately L = R where is in radians

s = R

ds/dt = R d/dt = R

Speed is ds/dt so v = R

A particle is moving around a circle with speed v:

sL

A baseball is thrown horizontally with an initial speed v. How far has it dropped after movinghorizontally through a distance L?

A. ½ g L/vB. ½ g v/LC. ½ g (L/v)2

D. ½ g (v/L)2

E. 2 g L/v

M

m

How large is m if system is in equilibrium?

A. m = MB. m = M / sin()C.m = M sin()D.m = M cos()E. M = M / cos()

vbox

Truck is moving with speed v and startsto decelerate with acceleration -a,Mass of truck is M, mass of box is m.

What force on the box is needed to keep the box stationary on the truck?

A. maB. 0C. mgD. (M+m)a

What is direction of the force on box needed to keep the box stationary on the truck?

A. forward B. backward

T2 =

A. m1aB. (m1+m2)aC. (m1+m2+m3)aD. 0E. (m1+m2-m3)a

What horizontal force is required to move the block at constant speed?

A.Mg / sin B.Mg cos C.Mg sin D.Mg tan E.Mg / cos

The mass is turned on its side socontact area is smaller.The force needed to overcome friction is:

A. SmallerB. LargerC. Same

A plane moves at constant speed v, in a horizontal circle of radius r.The plane is “banked” at an angle so that The total force (gravity + normal force) on a passenger of mass m, is upward from the seat.Does the correct bank angle depend on m?

A. YesB. No

A block with mass m, is on a horizontal surface having friction.The block has initial velocity v0

The block will eventually stop.Does the distance travelled depend on m?

A. YesB. No

The work done in moving the box up the ramp isA. Mg LB. Mg L Cos C.Mg L2

D.Mg L Sin E. ½ Mv2

Mass m1 is at rest.An equal mass m2, moves with velocity v, and collides with it. The collision is INELASTIC.What is the final velocity of the mass m2?

A. 0B. vC.v/2D.v/4E. v/Sqrt(2)

Mass m1 is at rest.An equal mass m2, moves with velocity v, and collides with it. The collision is ELASTIC.What is the final velocity of the mass m1?

A. 0B. vC.v/2D.v/4E. v/Sqrt(2)

Non-clicker question:

We assign to a dielectric (E=E0 /)

A conductor has what value of ?