TEST 2 Q 1 some HONORS review questions to try

Define:
displacement, velocity, average velocity, average speed, acceleration.

*Displacement:
change in distance from start (with direction)*

*Velocity:
change in displacement over change in time, (indicates direction)*

*Average
velocity: total change in displacement over change in time, the average
displacement for each time interval*

*Average
speed: the average distance traveled for each time interval, total distance
traveled over change in time.*

*Acceleration:
the increase in velocity over each time interval, change in velocity over
change in time.*

Describe the
motion of an object that starts with:

Positive
velocity, positive acceleration* it speeds up going forwards, displacement
always increasing. **à**. . . . . . . . .
.
. . **à*

* *

Positive
velocity, negative acceleration *it starts going forwards, getting slower,
but still going forwards, eventually stopping for a split second then reversing
direction and going backwards. **à**. . .
. . . . . . . . . **à*

* **ß**. . .
. . . . . . . . . **ß**↩*

* *

Negative
velocity, positive acceleration * it starts off going backwards fast, then gets slower while
still going backwards, eventually stopping for a split second then reversing
direction and going forwards.**ß**. . . . . . . . .
. .
. **ß*

* **➘**à**. . . . . . . . . . . . **à*

Negative
velocity, negative acceleration * It speeds up going backwards, displacement from start is
always decreasing (getting farther away in the reverse direction) **ß**. . . . . . . . . . .
. **ß*

When does an
object have an average velocity of zero?*When the displacement is zero (it
returns to its starting point)… note that its speed could be anything!*

** How do you
calculate total distance for an object that has a starting velocity of zero,
accelerates, then stays the same speed, then slows down to zero again, if:

all
three parts of the journey are the same time.

*Then
the total distance is the distance for each part added… or the average velocity
times the total time… In this case the average velocity for the first and last
part is ½ the final speed, if each time is equal, you can add up each average
velocity and divide by 3…. Vavg is 2/3 V final (1/3 (1/2V)+1/3(V)+1/3(1/2V))*

* D=2/3V*T*

All
three parts of the journey are the same distance.

*This
is harder to do, each distance is the same, yet each time is different, so need
to add the three distances….. D1=D2=D3 D1=V ^{2}/2A=D3*

A ball is rolled down a ramp.

Sketch the D-T and V-T graph.

*If down is positive*

Explain how to get the acceleration, using only a stopwatch and ruler. (T & D)

*If you know the initial velocity is zero, then the
average velocity is D/T. the final velocity is twice the average (middle)
velocity. Acceleration is change in velocity over time.*

* Or
using equation #4: A = 2D/T ^{2} if Vi=0*

If the a is 1 m/s^{2},
what is the ball’s velocity after 5 seconds? *Vi=0, A=1, T=5, Vf=???*

* Vf=Vi+AT
= 0+AT = AT = 1(5) = 5 m/s… the ball increases velocity by 1 m/s every second
for 5 seconds…*

What is its
average velocity? * If the
ball has a final velocity of 5 m/s, and an initial velocity of 0 m/s, its
average velocity is Vavg= (Vi+Vf)/2=(0+5)/2=5/2= 2.5 m/s*

How much
distance has it covered in that time? *Dis is Vavg*T = 2.5 (5)=12.5 meters*

* Or
D=ViT+ ½*AT ^{2} = 0+ ½(1)(5^{2})=1/2*25= 12.5 m*

What was the
distance traveled in the 5^{th} second? (from 4 to 5)

*The
distance traveled in the 5 ^{th} second is the distance at 5 sec minus
distance at 4 sec….*

*Distance
at 4 sec = D=ViT+ ½*AT ^{2} = 0+ ½(1)(4^{2})=1/2*16= 8 m, so
distance in the 5^{th} second is D(5)-D(4)= 12.5-8= 4.5 m*

*Or…. The
average velocity from T=4 to T=5 is (Vf(4)+Vf(5))/2= 4.5 m/s in one second…
D=Vavg*T=4.5*1=4.5 m*

* *

--------------------------------------------

A ball is
thrown up in the air and caught at the point of release.

Sketch the D-T
and V-T graph.

Explain what
two pieces of information would help calculate the maximum height.*You
know that at the maximum height the object has a Vf of 0, and the time is half
the total time of the trip.*

Explain what
would be different about the ball’s path, its maximum height, its graphs, and
its time and velocity if this were done on Jupiter (greater gravity).

*The
gravity is more on Jupiter, thus the ball would not go as high, and hit the
ground sooner, the acceleration due to gravity would be different, but the
final velocity at which it hits the ground would still be the same as the
start….*

* *

A speeder going
at a constant speed passes a police car, who starts accelerating from 0 some
time later. Explain how to calculate when and where the cops catch the speeder,
and sketch a strobe motion dot image, D-T, and V-T graph for each.

*If the
speeder is going at a constant speed, then his distance is rate times time… or
D=VT . . .
. . . .
. . . .
. . . . .
. . . .
.*

* *

*If the
police car is starting at Vi=0, then his distance is average velocity (1/2
final velocity=1/2AT) times time or D= 1/2AT ^{2}*

* .
. . .
. . . . .
. .
.
.*

* *

*They meet
at the same place (D) at the same time (T) D=D, T=T, so set the two equations
equal and solve for D and T…… VT=1/2AT ^{2}*

WORD PROBLEMS TO TRY (HONORS):

A tennis ball is released at the
top of a 5-m ramp and rolls down .

The ball reaches the end of the ramp
in 5.0 s and rolls onto the floor. If the ball experiences an

average deacceleration of -0.25
m/s ^{2 }as it rolls along the floor, how far from the end of the ramp
will the ball stop?

*Two parts, one on the ramp,
one after…….*

*On the ramp: Vi=0, D=5, T=5,
Vf=?, A=?*

* D=(Vi+Vf)/2
*T D=Vf/2*Tor
Vf=2Vavg=2*D/T=2*1=2 m/s*

*This Vf for part one is Vi
for part two:*

*Vi=2 m/s, A=-0.25, Vf=0
(when stops), D=????*

*For D, eq
#5 Vf ^{2}=Vi^{2}+2AD 0^{2}=2^{2}+2*-.25*D D= (0-4)/-.5=8 meters =D*

* *

I start traveling at
3 m/s and accelerate at the rate of 2 m/s^{2}, to a max of 25 m/s. I
can only last 5 seconds at this top speed though….

A tiger is 50
meters behind me traveling a constant 20 m/s.

When I reach my
maximum speed of 25 m/s, will I be in front of, or behind the tiger?

*Me: Vi=3
m/s, A = 2, Vf=25 m/s Dme=??? Tme=????*

* For
D, eq #5 Vf ^{2}=Vi^{2}+2AD 25^{2}=3^{2}+2*2*D D= (625-9)/4=154 meters =Dme*

* *

* For
T, eq #3 Vf=Vi+AT 25=3+2(T) T=(25-3)/2=11
seconds*

* *

*In that
time…… Tiger Di =-50, Vi=Vf=Vavg=20m/s, A=0*

* D=20T-50 Dtiger=20(11)-50=220-50=170m*

*The tiger
is still ahead of me!!!*

Will I
eventually ever be ahead of him?

*Since I
can only keep up this pace for another 5 seconds*

*Dme=154+VT=154+25(5)=279m*

*Dtiger=170+20(5)=270m so
eventually I will be ahead of him…. Of course, then I will stop and he will
pounce on me and that’s it for me **L**.*

* *

I stand on a
cliff and jump up at 39.2 m/s. If I hit the valley below after 10 seconds, how
fast am I going? How high is the cliff? How high did I jump?

*Vi=39.2
m/s, A=-9.8 m/s ^{2}, T=10sec*

* Vf=??,
D=???*

* Vf=
Vi +AT = 39.2 + -9.8 * (10) = -58.8 m/s downward*

* D=ViT+1/2AT ^{2}
= 39.2 (10) + ½*-9.8*100 = -98 meters below start is cliff*

* *

*How high?
At the apex Vf = 0 m/s so:*

* Vi=39.2
m/s, Vf=0m/s, A=-9.8 m/s ^{2}, Dheight=????*

*For D, *

*eq #5 Vf ^{2}=Vi^{2}+2AD 0^{2}=39.2^{2}+2*-9.8*D D= (-1536.64)/-19.6=78.4 meters
=Dheight*

If my friend
saw me and started running to catch me, with an acceleration of 3 m/s^{2},
how close would he have to be to the cliff to catch me?

*If my
time was 10 seconds, his vi=0, and A=3,he would need to be *

*D=ViT+1/2AT ^{2}
= 0 (10) + ½*3*100 = 150 meters away from the cliff to catch me.*