**PHY
1151**

**Doug
Davis**

10, 12, 16, 19, 23, 38, 46, 48, 51, 62, 66, 86, 92, 96, 101, 102, 106

| **ToC,
Chapter 2**** |
****Course
Calendar** |

**Everything** will come from
our "Big Three Kinetmatics Equations":

x = x_{i} + v_{i}
t + (^{1}/_{2}) a t^{2}

v^{2} =
v_{i}^{2} + 2 a (x - x_{i})

**2.10** A bicyclist travels
with an average velocity of 15 km/h, North for 20 minutes. What is
his displacement?

The definition of average velocity, v = can be solved for Ds = v Dt

**2.12** The speed of sound in
air is about 330 m/s. You observe a lightning bolt strike a tree 1.5
km away. How much time will elapse between your seeing the lightning
bolt and hearing the thunder that accompanies it? Why do you not need
to take into account the speed of light?

The definition of average velocity, v = can be solved for Dt =

**2.16** A car accelerates from
rest to 90 km/h in 8.8 s. What is its average acceleration in
m/s^{2}?

Use the basic definition of acceleration, a =

It may be easier, or more convenient, to change the velocity of 90 km/hr into units of m/s early on,

a =
= =
**2.8 m/s ^{2}**

**2.19** A jet aircraft landing
on an aircraft carrier is brought to a complete stop (v_{f} =
0) from a (an initial) velocity of (v_{i}= ) 215 km/h in 2.7
seconds. What is its average acceleration in
m/s^{2}?

Again, use the basic definition of acceleration, a =

As before, it may be easier, or more convenient, to change the velocity of 90 km/hr into units of m/s early on,

v_{f} = 0

a =
= =
= -
22.1 m/s^{2}

**2.23 ** A certain car has an
acceleration of 2.4 m/s^{2}. Assume that its acceleration
remains constant. Starting from rest, how long does the car require
to reach a velocity of 90 km/h? How far does it travel while reaching
that velocity?

It may be easier, or more convenient, to change the velocity of 90 km/hr into units of m/s early on,

We can solve the equation v =
v_{i} + a t for the time t, t =
.

"Starting from rest" means
v_{i} = 0 so that becomes t =

Now that we know "how long" we can use the displacement-time equation, with this time, to determine "how far".

s = 0 + 0 (10.4 s) +
(2.4 m/s2) (10.4 s)^{2}

**s = 128
m**

**2.38** In passing a slow
truck you must accelerate from 50 km/h to 90 km/h. What must your
acceleration be so you can complete your passing maneuver in 1.0
km?

We know the velocities and the displacement involved and don't really care about the time so we can use the equation

It may be easier, or more convenient, to change the velocities of 50 km/hr and 90 km/hr into units of m/s early on,

v_{f} = 90
( ) (
) (
) = 25

v^{2} =
v_{i}^{2} + 2 a (s - s_{i})

( 25
)^{2} = ( 13.9
)^{2} + 2 a ( 1000 m )

625 m^{2}/s^{2} =
193 m^{2}/s^{2} + (2000 m) a

(2000 m) a = (625 - 193)
m^{2}/s^{2} = 432
m^{2}/s^{2}

a =
= **0.216 m/s ^{2}**

**2.46** A car, speeding at 130
km/h passes a police car at rest. Just as the speeding motorist
passes the police car, the police car begins pursuit. If the police
car maintains a constant acceleration of 5.8 m/s^{2}, when
and where will the speeding motorist be overtaken?

We can make life a little easier by changing the motorist's velocity of 130 km/hr into units of m/s.

The motorist travels at
__constant__ velocity of 130 km/hr or 36.1 m/s so the motorist's
acceleration is zero, a_{mot} = 0 .

The motorist's position or displacement is given by

s_{mot} = 0 + (36.1 m/s) t
+
(0) t^{2}

s_{mot} = (36.1 m/s)
t

The police car starts from rest so
v_{pol, i} = 0 but then accelerates at 5.8
m/s^{2}.

The police car's displacement or position is given by

s_{pol} = 0 + (0) t +
(5.8
m/s^{2}) t^{2}

s_{pol} =
(5.8 m/s^{2}) t^{2}

For what **time** and
**where** are these two displacements equal?

The motorist's position or displacement is given by

The police car's displacement or position is given by

For what **time** and
**where** are these two displacements equal? Set the two equations
equal to each other.

(5.8 m/s^{2}) t^{2} = (36.1 m/s) t

(5.8 m/s^{2}) t = (36.1 m/s)

t = s = 12.5

**t = 12.5
s**

That tells us **when**. We can
put that back into either displacement equation to determine
**where**. The motorist's displacement, since it has only a term
with time and not with time-squared, is the easier to use.

And we can __check__ this
answer by evaluating the police car's displacement.

The answers agree to **two**
**sig**nificant **fig**ure**s**. The time of 12.5 s was
really 12.448 s and we would expect a closer agreement between the
two answers if we kept more significant figures in our answer for the
time. But more significant figures are __not__ warranted because
the police car's acceleration of 5.8 m/s^{2} is given only to
**two** **sig**nificant **fig**ure**s**.

**2.48 **A late commuter,
sprinting at 8 m/s, is 30 m away from the rear door of a commuter
train when it starts to pull out of the station with an acceleration
of 1 m/s^{2}. Can the commuter catch the train (if the
platform is long enough)?

This time, notice that the
commuter and the train **start from different places**. That is,
they have different initial displacements. We can take care of that
with

The initial velocities of commuter and train are

Their two accelerations are

We can write equations for the displacement of the commuter and for the displacement of the train

s_{c} = 0 + (8 m/s) t +
(0) t^{2}; and s_{t} = 30 m + (0) t +
(1 m/s^{2}) t^{2}

s_{c} = (8 m/s) t; and
s_{t} = 30 m +
(1 m/s^{2}) t^{2}

Now set these two equations equal
to each other and solve for the time t. If there is __no__
solution, then the commuter can not catch the train. If there is a
solution, that value of time t tells us __when__ the commuter
catches the train. We can then use that value of time t to evaluate
the displacement from either equation.

30 m +
(1 m/s^{2}) t^{2} - (8 m/s) t = 0

(1 m/s^{2}) t^{2} - (8 m/s) t + 30 m = 0

(1 m/s^{2}) t^{2}
- (16 m/s) t + 60 m = 0

If time t is measured in
__seconds__, then each term in this equation has units of m (for
meters) and we may eliminate the units and simply write

This is now a quadratic equation of the form

and has a solution in the form of

For our present equation, we have a = 1, b = - 16, and c = 60.

Therefore,

t =

t =

t =

t =

There are ** two**
solutions:

and

What do these ** two**
solutions mean?

At t = 6 seconds, the commuter
catches up with the train. This is the "real solution". At t = 6
seconds, the commuter catches the train and the race is over.
However, the mathematics does not stop there. If the commuter
continues to run, he will __pass__ the train and the train will
catch up with him at t = 10 seconds!

**2.51** While standing still,
a police car is passed by a speeding car traveling at a constant 120
km/h. The policeman waits 2 seconds before deciding to pursue. What
must be his acceleration in order to catch the speeding car within
three kilometers?

First, let's do the easy conversion of the car's speed of 120 km/hr to m/s

Remember, since the car travels at
constant velocity, this is its __initial__ velocity and we know
its acceleration is zero, a_{car} = 0

The police car's initial velocity
is zero, v_{pol, i} = 0 and we are trying to solve for its
acceleration, a_{pol} = ?

We can immediately write the equation for the displacement or position of the car,

s_{car} = 0 + (33.3
) t +
(0)
t^{2}

s_{car} = (33.3 )
t

We can set this equal to 3 km or 3,000 m and find out how long the car travels before the police car catches up with it.

t = = 90 s

(3 km is a l-o-n-g distance for catching the speeding motorist).

Remember, the policeman
__waits__ for 2 s before beginning pursuit. Therefore, the time
required for the police car to travel 3,000 m is **88 s**. We can
write an equation for the displacement of the police car and find the
acceleration necessary.

3,000 m = 0 + (0) (88 s) +
apol (88 s)^{2}

3,000 m =
a_{pol} (88 s)^{2}

3,000 m = (3872 s^{2})
a_{pol}

a_{pol} =

**a _{pol} = 0.77
m/s^{2}**

**2.62** How fast must a stone
be thrown, straight up, to just reach the top of a building that is
10 m high?

Throughout the stone's flight, its
acceleration is a = - 9.8 m/s^{2} (for convenience, we will
use the approximation that its acceleration is a = - 10
m/s^{2}). This means we have taken **up** as
**positive**.

Since we do not care about the time, we can use the third of our "big three" equations,

or

or

We will measure everything from
the ground so s_{i} = 0. At the top, where s = 10 m, the
stone stops momentarily so v = 0.

0^{2} =
v_{i}^{2} + 2 ( - 10 m/s^{2}) ( 10 m -
0)

0 = v_{i}^{2} -
200 m^{2}/s^{2}

v_{i}^{2} = 200
m^{2}/s^{2}

**v _{i} = 14.1
m/s**

**2.66** Cliff divers in
Acapulco dive from rocks about 35 m above the water. Neglecting air
resistance, what is their velocity as they hit the water?

Everything is happening in the
"down" direction, so we may as well take **down**
as **positive** for this case. If
down is positive, then the acceleration is a = + 10 m/s^{2}
(the acceleration is really + 9.8 m/s2, but we will make the
arithmetic easier by using this approximation). We will measure
distances __from__ the top of the cliff.

There, at the top of the cliff,

Since we are not interested in the time, we can use the third of our "big three" equations,

v^{2} = 0^{2} + 2
(10 m/s^{2}) ( 35 m - 0 )

v^{2} = 700
m^{2}/s^{2}

**v = 26.5
m/s**

**2.86** In the mid-1960’s
McGill University, in Montreal, launched high-altitude weather
sensors by firing them from a cannon made from two World War II Navy
cannon bolted together for a total length of 18 m. It was proposed
that they use this arrangement to launch a satellite. Orbital speed
of a satellite is about 29 000 km/h. What would have been the average
acceleration throughout the 18 m length of the cannon to have a
muzzle velocity of 29 000 km/h?

While it is easier to __think__
of orbital velocity in terms of km/h, it is easier to do the
calculations with this velocity in units of m/s.

v^{2} =
v_{i}^{2} + 2 a ( s - s_{i})

(8060 )^{2}
= 0^{2} + 2 a ( 18 m - 0)

(8060 )^{2}
= 02 + 2 a ( 18 m - 0)

64 963 600 = (36 m) a

a =

a = 1 804 544
m/s^{2}

**a = 1 800 00
m/s ^{2}**

We will later see that there is an
important connection between forces and accelerations. To experience
such an __enormous__ acceleration, the payload would have to be
able to withstand __enormous__ forces!

**2.92** A falling flower pot
takes 0.25 s to pass by a window 1.5 m high. How far above the top of
the window was the balcony from which the flower pot fell?

(Only)
__After__ the diagram is carefully drawn and labeled, we can start
writing equations (notice that I have taken __down__ as
__positive__):

We are looking for the distance I
have labeled **y _{top}**, the distance from the balcony
above to the top of the window.

y_{top} = y_{i } +
v_{yi} t_{top} + (^{1}/_{2}) a
t_{top}^{2}

y_{bot} = y_{i} +
v_{yi} t_{bot} + (^{1}/_{2}) a
t_{bot}^{2}

Since we are measuring distances
from the balcony above, we know y_{i} = 0. Since the flower
pot falls from rest, we know v_{yi} = 0. Then our equations
for y_{top} and y_{bot} can be simplified to become
just

y_{bot} =
(^{1}/_{2}) (9.8 m/s^{2})
t_{bot}^{2}

But we also know y_{bot} =
y_{top} + 1.5 m and t_{bot} = t_{top} + 0.25
s.

y_{top} = (4.9
m/s^{2}) t_{top}^{2}

y_{bot} =
(^{1}/_{2}) (9.8 m/s^{2})
t_{bot}^{2}

y_{top} + 1.5 m = (4.9
m/s^{2}) (t_{top} + 0.25
s)^{2}

Subtract one equation from the other,

- [ y_{top} = (4.9
m/s^{2}) t_{top}^{2} ]

1.5 m = (4.9 m/s^{2}) (
[0.5 s] t + 0.0625 s^{2})

1.5 m = (2.45 m/s) t + 0.306 m

1.194 m = (2.45 m/s) t

(2.45 m/s) t = 1.194 m

t = [ 1.194 m ] / [2.45 m/s ]

t = 0.487 s

This is really t_{top} but
I dropped the subscript to make things look "prettier" (ie, less
messy or less busy).

But we want the __distance__ to
the top,

y_{top} =
(^{1}/_{2}) (9.8 m/s^{2})
t_{top}^{2}

y_{top} = (4.9
m/s^{2}) t_{top}^{2}

y_{top} = (4.9
m/s^{2}) (0.487 s)^{2}

y_{top} = 1.163
m

**y _{top}** = 1.16
m

**2.96** Poor Wyl E Coyote is
at it again. Attempting to catch the Road Runner he falls from the
top of a 400 m cliff. He is wearing an Acme Fireworks rocket but
requires 6 seconds to light the fuse and ignite the rocket. The
burning rocket gives him an upward acceleration so that his velocity
just reaches zero as he reaches the bottom.

a) How far has he fallen before the rocket starts?

b) What is his upward acceleration in order to have this gentle landing?

But, alas, as Edsel Murphy would have it, the Acme rocket does not shut off. Nor is the Coyote able to release himself from the rocket. Therefore, the rocket now carries him upward with this same acceleration for an additional 5 seconds before it depletes its fuel.

c) How high is he then (as the fuel runs out)?

d) How fast is he moving then?

e) How high -- how far above the canyon floor -- does the Coyote eventually go?

f) How fast is he going as he hits the canyon floor this time?

In t = 6 s, the Coyote has fallen a distance s (or y) given by

or

We will measure displacements
__from the cliff__ so that s_{i} = y_{i} =
0

At time t = 6 s, this becomes

y =
(- 10 m/s^{2}) (36 s^{2})

**y = - 180
m**

**a) y = - 180 m**, that is,
**180 m below the top of the cliff** (or 220

The burning rocket gives him an upward acceleration so that his velocity just reaches zero as he reaches the bottom.

b) What is his upward acceleration in order to have this gentle landing?

Be careful. It might be better to write this equation with subscripts 1 and 2.

Now v_{1} and
s_{1} refer to the situation we found at the end of part
**a)**,

And v_{2} and
s_{2} refer to the velocity and position right at the canyon
floor,

v_{2}^{2} =
v_{1}^{2} + 2 a (s_{2} -
s_{1})

(0)^{2} = (- 60
m/s)^{2} + 2 a [(- 400 m) - (-180 m)]

0 = 3600
m^{2}/s^{2} + 2 a (- 400 m + 180 m)

0 = 3600
m^{2}/s^{2} + 2 a (- 220 m)

2 a (220 m) = 3600
m^{2}/s^{2}

(440 m) a = 3600
m^{2}/s^{2}

a =
= 8.18 m/s^{2}

**a = 8.18
m/s ^{2}**

**c) **

From the floor of the canyon, from
being at rest, Wyl E Coyote accelerates __upward__ with an
acceleration of 8.18 m/s^{2}. We __could__ start
everything over at this point and measure distances from the floor of
the canyon. However, it is just as good -- and just as easy -- to
leave the origin where it was and continue to measure distances from
the top of the cliff. With that reference frame, the initial position
for this part of the trip is

y = y_{i} +
v_{y,i} t +
a_{y} t^{2}

y = (- 400 m) + (0) (5 s) +
(8.18
m/s^{2}) ( 25 s^{2})

y = - 400 m + 103 m

**y = - 297
m**

That is, 297 m __below__ the
cliff (or 103 m above the canyon floor).

**d)** How fast is he moving
then?

v = 0 + (8.18 m/s^{2}) (5
s)

**v = 40.9
m/s**

Note that the __positive__
value for the velocity means the Coyote is moving __up__ when the
fuel runs out.

Now the Coyote is in "free fall"
with a = - g = - 10 m/s^{2} (yes, his acceleration is
__really__ a = - g = - 9.8 m/s^{2} but we will again use
this approximation just to keep the arithmetic somewhat easier and to
be consistent with the earlier use of it).

The equation connecting velocity, acceleration, and distance is

or

where s_{1} =
y_{1} = - 297 m, v_{y1} = v_{1} = 40.9 m/s,
and v_{y2} = v_{2} = 0 and we are wanting to
__find__ s_{2} or y_{2}.

0 = (40.9 m/s)^{2} - (20
m/s^{2}) [y_{2} + 297 m]

0 = 1673
m^{2}/s^{2} - (20 m/s^{2}) y_{2} -
5940 m^{2}/s^{2}

0 = 1673
m^{2}/s^{2} - (20 m/s^{2}) y_{2} -
5940 m^{2}/s^{2}

0 = - (20 m/s^{2})
y_{2} - 4267 m^{2}/s^{2}

(20 m/s^{2}) y_{2}
= - 4267 m^{2}/s^{2}

y_{2} = -

**y _{2} = - 213
m**

Remember, that is 213 m
__below__ the cliff or 187 m __above__ the canyon
floor.

**f)** Alas, the rocket fuel is
finally depleted and our Coyote falls this 187 m to the canyon floor
below. What is his impact velocity?

Again, since we are not particularly interested in the time we can use

or

v_{y2}^{2} =
(0)^{2} + 2 (- 10 m/s^{2}) [(- 400 m) - (- 213
m)]

v_{y2}^{2} =
(0)^{2} + 2 (- 10 m/s^{2}) (- 187 m)

v_{y2}^{2} = 3740
m^{2}/s^{2}

v_{y2} = ± 61.2
m/s

__Either__ + 61.2 m/s or - 61.2
m/s is a solution to the __mathematics__. What is happening
__physically__? The Coyote is falling __down__ so we must
choose the __negative__ solution,

**2.101** The figure below is a
position-time graph for a lab cart traveling along a straight track
in a Physics lab. Determine the time(s) that the cart

a) has its greatest velocity.b) has constant velocity.

c) is moving backward.

**a)** has its greatest
velocity. Velocity is the __slope__ of the line and that seems to
be greatest around **7.5 seconds** or so.

**b)** has constant velocity.
This means constant __slope__. From about **4.5 s** to **5.5
s** the slope and velocity are constant at **zero**. From about
**1.5 s** to about **3.0 s** the slope and velocity seem to be
constant with a value of about 0.4 m/s. From about **8.5 s** on
the slope and velocity seem to be constant with a value of about -
0.6 m/s

**c)** is moving backward.
"Moving backward" means a __negative__ slope. That occurs between
**3.5 s** and **4.5 s**, between **5.75 s** and **6.5
s**, and from **8.25 s** on to the end of the graph around **11
s**.

** **

**2.102** The Figure below is a
position-time graph. From it construct the corresponding
velocity-time and acceleration-time graph. (That is, do not worry
about the acceleration-time graph as the data given is really not
sufficient).

The velocity is the __slope__
of the line on a position-time graph. From about 1.75 s to about 3.5
s the position increases at a rate of 1.0 m/s; that is, the velocity
is 1.0 m/s during that time. Then the velocity starts to decrease,
passing through zero at about 4.0 s. By about 4.25 s, the position is
decreasing at a steady rate of 0.5 m/s so the velocity is a constant
- 0.5 m/s up until about t = 8.0 s. Then the velocity approaches zero
and remains at zero from about 8.25 seconds on.

There is really insufficient information to accurately find the acceleration.

**2.106
****--
Oops! I seem to have 2.106 on here twice. Take your pick. I
hope the two solutions are consistent**.--

**2.106** The Figure below is
an acceleration-time graph. Use it to construct corresponding
position-time and velocity-time graphs. Start the car from rest, vi =
0, at x = 0 for t = 0.

During the time from 0 s to 2.0 s, the acceleration is a = + 1 m/s2.

v = v_{i} + a t

v(2 s) = 0 + (1 m/s^{2}) (
2 s) = 2 m/s

From t = 2 s to t = 6 s, the acceleration is zero

so the velocity remains
__constant__ at v = 2 m/s.

During the time from 6.0 s to 8.0
s, the acceleration is a = - 1 m/s^{2}.

v(8 s) = 2 m/s + (- 1
m/s^{2}) ( 2 s) = 0

From t = 8 s to t = 12 s, the acceleration is zero

so the velocity remains
__constant__ at v = 0.

Now, for the position or displacement.

During the time from 0 s to 2.0 s,
the acceleration is a = + 1 m/s^{2}.

s (2 s) = 0 + 0 (2 s) +
( 1 m/s^{2}) (2 s)^{2}

s (2 s) = 2 m

It is __only__ a
__coincidence__ that the numerical value of the displacement
happens to be equal to the numerical value of the velocity for time t
= 2 s.

From t = 0 to t = 2, the
position-time curve is a __parabola__ that goes through (0, 0) and
(2, 2). From t = 2s to t = 6 s, the velocity remains constant at v =
2.0 m/s so the position-time curve is a __straight__ __line__
with a slope of 2.0 m/s. This means the displacement for t = 6 s
is

Dt = 4
s since this piece of the problem starts at t_{i} = 2
s.

From t = 6 s to t = 8 s, the acceleration is - 1.0 m/s2, with the velocity going from 2.0 m/s to zero during that time. At t = 6 s, which we will call ti = 0, the displacement is 10 m. We can calculate the displacement at the end of this Dt = 2 s interval by

s = 10 m + (2 m/s) (2 s) +
(- 1
m/s^{2}) (2 s)^{2}

s = ( 10 + 4 - 2 ) m = 12 m

We can put all these together on a graph. The circles on the graph above indicate the displacements we have actually calculated. The curve from 0 to 2 s is a parabola. The line from 2 s to 6 s is a straight line. The curve from 6 s to 8 s is a parabola. And the line from 8 s on is a straight, horizontal line, corresponding to zero velocity.

** **

**2.106** **(second
version!)** The figure below
is an acceleration-time graph. Use it to construct the corresponding
position-time and velocity-time graphs. Start the car (or whatever)
from rest, vi = 0, at x = 0 for t = 0.

First, look at the motion for time
between 0 and 2.0 s (0 < t < 2.0 s) for which the acceleration
is a = 1.0 m/s^{2}. We can calculate the velocity
from

v = 0 + (1 m/s^{2})
t

v = (1) t

At the end of that 2.0 seconds (at t = 2.0 s), the velocity is 2 m/s. That appears on a velocity-time graph as

From2.0 to 6.0 seconds-that is, for 2.0 s < t < 6.0 s - the acceleration is zero ( a = 0 ) which just means the velocity remains constant ( v = 2.0 m/s). That appears on the velocity-time graph as a horizontal line.

From 6.0 to 8.0 s - that is, for 6.0 s < t < 8.0 s - the acceleration is - 1.0 m/s2 . Then the velocity is given by

v = 2.0 m/s + ( - 1.0
m/s^{2}) t

v = 2.0 m/s - (1.0
m/s^{2}) t

That is a straight line starting
at v = 2.0 m/s and having a slope of - 1.0 m/s^{2}. That part
of the velocity-time graph looks like this:

For the rest of the time, 8.0 s
< t < 12.0 s, the acceleration is again zero so the velocity
remains constant. During the two seconds from t = 6.0 s to 8.0 s, the
velocity has been reduced from 2.0 m/s back to 0.0 so the object is
at rest at 8.0 s. It __remains at rest__ for the rest of the
time.

Now, we will construct
**position**-time graphs. As with the previous part, we will look
at each time segment that has a constant acceleration. For 0 < t
< 2.0 s, the acceleration was 1.0 m/s2 and the initial velocity
was zero. Our position equation is

x = 0 + 0 t +
(^{1}/_{2}) (1 m/s^{2})
t^{2}

x = (0.5 m/s^{2})
t^{2}

This is a curve-a parabola,
actually-that starts at (x = 0, t = 0) and ends at (x = 2 m, t = 2
s). On a graph that looks __something__ like this:

From 2.0 s to 6.0 s, the velocity
remains __constant__ at v = 2.0 m/s. This means the position,
which is x = 2.0 m at time t = 2.0 s, increases at a steady, linear,
constant rate.

x = 2.0 m + (2.0 m/s) t +
(^{1}/_{2}) (0) t^{2}

x = 2.0 m + (2.0 m/s) t

We started this segment, at
t_{clock} = 2.0 s, at a position of 2.0 m. At the end of the
segment, with t_{clock} = 6.0 s, we are at position x = 10
m.

On a graph, that looks like this:

Now, at t = 6.0 s, there is a
negative acceleration. At t = 6.0s, the velocity is 2.0 s; that
becomes vi for this time segment. At t = 6.0 s, the position is 6.0
m; that becomes xi for this segment. For 6.0 s < t_{clock}
< 8.0, we have a = - 1.0 m/s2 and the displacement equation
becomes

x = 10.0 m + (2.0 m/s) t +
(^{1}/_{2}) (- 1.0 m/s^{2})
t^{2}

x = 10.0 m + (2.0 m/s) t - (0.5
m/s^{2}) t^{2}

At the end of this segment, for t = 8.0 s, the position (or displacement) is

[Be careful. I have written tclock = 8.0 s to keep track of our time overall. But our equations have all been developed with ti = 0 corresponding to xi and v, so the time that goes into this equation must be 2.0 s. If this is confusing, please let me know].

x (t_{clock} = 8.0 s) =
12.0 m

This means a curve on the graph that looks something like this:

Now, for the final segment, with
8.0 s < t_{clock} < 12.0 s, the object remains at rest
(v = 0) so the displacement remains constant, x = 12.0 m.

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(c) 2002, Doug Davis; all rights reserved