Graph this (assume motion along a line)You walk to the bus stop in the morning (400 meters away) in 4 minutes. After 1 minute, you realize that you forgot your clicker. You return to your room quickly, in 2 minutes, but it takes you 1 minute to find your clicker. You return to the bus stop in 2 minutes just in time. x [m]

t [min]

400

00 4 82 6 10

Next: what is the velocity in each interval(that’s the slope in each section)?

€

v =Δx

Δt

Now draw the velocity versus time graph

Recall that velocity is

This can be interpreted as the slope of the x versus t graph

€

v =Δx

Δt

v [meter/min]

t [min]

200

00 4 82 6 10

-200

100

-100

Here is a motion diagram of a car moving along a straight stretch of road.

If the positive-x direction is to the right, which of the following velocity-versus-time graphs matches this motion diagram?

Checking Understanding

A. B. C. D.t t t t

v v v v

4

Position and Velocity vs. TimeA particle moves with the velocity-versus-time graph shown.

Which graph best illustrates the position of the particle as a function of time?

A. B. C. D.

5

Position and Velocity vs. TimeA particle moves with the velocity-versus-time graph shown.

Which graph best illustrates the position of the particle as a function of time?

A. B. C. D.

6

AccelerationThese three motion diagrams show the motion of a particle along the x-axis. Rank the accelerations corresponding to these motion diagrams, from most positive to most negative. There may be ties.

1

2

3

x

7

AccelerationThese three motion diagrams show the motion of a particle along the x-axis. Which has positive acceleration? Which is negative?

1

2

3

1 < 3 < 2

x

Free fallConstant downward acceleration:

a = –g = –9.8 m/s2

8€

v(t) = v i − g(t − ti)

y(t) = y i + v i (t − ti) −g

2(t − ti)

2

v f2 − v i

2 = −2g Δy

Free Fall Example with stoppingPassengers on The Giant Drop, a free-fall ride at Six Flags Great America, sit in cars that are raised to the top of a tower. The cars are then released for 2.0 s of free fall. A) How fast are the passengers moving at the end of this speeding up phase of the ride (you can use g=-10m/s2)? B) If the cars in which they ride then come to rest in a time of 1.0 s, what is acceleration (magnitude and direction) of this slowing down phase of the ride? C) Given these numbers, what is the minimum possible height of the tower?0

vt[s]

tf1=ti2=2.0

vf1

3.0=tf20

a1=-9.8 m/s2a2

The graph helpsto organize theinformation givenin the problem.

Free Fall Example with stopping

B) During the stopping portion of the motion, we can use the same equation to find the unknown acceleration

€

v f = v i + (−g) (t f − ti)

v f = (0 m/s) + (−10 m/s2)(2.0 s) = −20 m/s

A) During the free falling portion of the motion, the acceler-ation is -g=-9.8 m/s2:

€

v f = v i + a2 (t f − ti)

0 m/s = (−20 m/s) + a2(1.0 s)

a2 = +20 m/s2

Note that the acceleration during the stopping portion is positive (same sign as the slope on the v versus t graph for this part of the motion)

Free Fall Example with stopping

Since the ride falls 30 meter, the tower had better be at least that high (about 100 feet)

C) To find the net displacement during the two motions, we need to separately find the displacements during each part

€

Δy = v i(t f − ti) +a

2(t f − ti)

2

Δy1 = (0 m/s)(2.0 s - 0 s) +−10 m/s2

2(2.0 s - 0 s)2 = −20 m

Δy2 = (−20 m/s)(3.0 s - 2.0 s) +20 m/s2

2(3.0 s - 2.0 s)2 = −10 m

Δynet = Δy1 + Δy2 = (−20 m) +(−10 m) = −30 m

y (m)

t (s)2 30

0

-20

-30