Friday 18 Sep:
Pre-Test Ch 1-3
(You can’t take the Real Test unless you pass this one with a minimum score on this one)
Show all work: No work à you won’t receive full credit.
- Sunlight (V = 3x108 m/s) takes 8.3 minutes to reach the Earth. How many meters is the sun away from us?
2.
2. Using Pos. vs. Time graph above, solve the following
- Velocity between 0-2 sec
- Velocity between 2-4 sec
- Avg. Velocity between 0-7 seconds
3.
3. Using Vel. vs. Time graph above, solve the following
- Acceleration between 3-4 sec
- What times was the object’s velocity = 0
- At what times was the object’s acceleration = 0
- Avg. Acceleration between 0-7 seconds
- Total distance traveled between 0 -7 seconds
- Displacement traveled between 0-7 seconds
4. You fire a bullet (1,600 m/s) at a bell and hear the ring 0.731 sec later (speed of sound is 330 m/s), how far away is the target?
5. A rabbit starts from rest and runs 20m in10 sec. Find:
a. Avg. Acceleration b. Final velocity
6. A row-boat is moving at 2.2 m/s. The rower stops rowing and coast to a stop in 10.0m. Find:
a. The boat’s acceleration? b. Time it took to come to a stop
c. How far did it travel during the 3rd sec of drift?
7.
North Bank
300
A speed-boat crosses the river from North Bank to South Bank at a speed of 10m/s.
River Current 10m/s
a. How fast must the river current be flowing to cause (result in) in a course directly south across the river?
South bank
b. The river is crossed in 150sec. How wide is the river?
8. An arrow is shot at a 53.10 angle from the horizon at 40m/s. Find
a. total time in air b. speed at top of arc
c. max height d. total horizontal dist e. speed when it hits the ground
9.
Cannon
cliff
300
100m/s
50m
Find:
a. Time
it takes projectile
to hit the ground. ground
b. Distance from base of cliff to where cannon ball strikes the ground.
c. Angle (measured from the ground) that the cannon ball (projectile) hits the ground.
10. How many seconds does it take the hour hand of a clock to make one complete cycle?
11. a. You swing an object on a string around in a horizontal circle (CCW) above your
head. Name the three forces and their direction acting on the
object.
b. Can the string be truly horizontal? 1 2 3
c. Why/why not? 4
d. If the string breaks while the object is in object
moving in the horizontal circle heading due
north at the moment of cut, what direction is the object’s path?
12. A baseball catcher throws a ball vertically upward and catches it in the same spot when it returns to his mitt. At what point in the ball’s path does it experience zero acceleration?
a. midway on the way up
b. at the top of its trajectory
c. the instant it leaves the catcher’s hand
d. the instant before it arrives in the catcher’s mitt
e. none, it always will experience an acceleration
13. A ball is thrown straight upward from the ground and lands 10.0 s later. What is the initial speed of the ball?
a. 50 m/s
b. 25 m/s
c. 10 m/s
d. 5 m/s
e. None of these
14. A car starts from rest and accelerates at 3.00 m/s2 through a distance of 60.0m. How long did it take the car to cover the 60.0 m distance?
a. 20.0 s
b. 6.32 s
c. 7.75 s
d. 5.16 s
Points for questions as shown below:
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Answers:
- x = vt à (3x108)(498s) = 1.49 x1011 cm
- a. -2/2 = -1.0 cm/s
b. 3/2 = 1.5 cm/s
c. 0/7 = 0 cm/s
- a. -2/1 = -2.0 cm/s
b. 0.0 s & 4.0 sec
c. 2.0 to 3.0s & at 5.0s
d. -1/7 = -0.143
e. ½( 2*2) + (2*2) + ½( 1*2) + ½( 1*2) + ½( 2*1) + (2*2) = 9.0 cm
f. ½( 2*2) + (2*2) + ½( 1*2) - {½( 1*2) + ½( 2*1) + (2*1)} = +1.0 cm
-
- Ratio Method:
330 / 1600 = .206 or approx 1/5 ratio
Therefore;
bullet took 1/6 of total time and sound took 5/6 of total time
(1/6)(.731sec) = .122s
(5/6) (.731s) = .609s
Bullet: x =vt à x = (1600m/s)(.122s) = 195m
Sound: x =vt à x = (330m/s)(.609s) = 201m
Avg = 200M
Alternative Method (equation method for math people)
Bullet: Vb = x/t à 1,600 =x/tb à tb = x/1600
Sound: Vs = x/t à 330 = x/ ts à ts = x/330
x = dist that bullet traveled and what sound traveled (sum of total dist)
tb = .000625x ts = .00303x
(Total time) 0.731s = tb + ts = .000625x + .00303x à .00366x
x = 0.731/.00366 à 200m
5. a. x =vit + 1/2at2 à 20 = 0 + ½ a (102) à a = 0.4 m/s2
- vf = vi + at à (.4)(10) = 4.0m/s
6. a. vf2 - vi2 = 2aDx à 0 – (2.22) = 2a(10) à
a = -0.242m/s2
b. vf = vi + at à 0 = (2.2) + (-0.242)t à
t = 9.09s
c. vf = vi + at à = 2.2 + (-0.242)(2) =2.2 - 0.484 = 1.72m/s (at the start of 3rd s)
x =vit + 1/2at2 à x =(1.72)(1) + ½(-.242)12 à x = 1.60 m
7. a. Cos300 (Hyp) = current speedà .867(10) = 8.67 m/s
- d = vt à (Sin300)(10m/s)(150s) = 750 M
8. a. vf = vi + at à -(Sin53.10)(40m/s) = (Sin53.10)(40m/s) + (-10)(t)
-32 – (32) / - 10 = t (in air) = 6.4s
b. speed at top = (Cos53.10)(40) = 24.0m/s
c. height: x =vit + ½at2 à 0 + ½ (10)(3.22) = 51.2m
d. Horz dist = (vhorz)(t) à (Cos53.10)(40m/s)(6.4s) = 148.8m
e. Final speed = Init speed à 40m/s
9. a. vf2 - vi2 = 2aDx (V = vertical V)
vf2– (Sin300)(100m/s) 2 = 2(-10)(-50) à vf2– 2500 = 1000 à √3500 = -59.2 m/s
vf = vi + at à (-59.2) = (-50) + (-10)t à -9.2/-10 = 0.92 sec
b. Horz dist = (vhorz)(t) à Cos300(Hyp)t à (.867)(100)( 0.92)= 79.8 m
c. ∟(Angle): Tan =vert V/ Horz V à Tan-1(59.2m/s / 86.7 m/s) = 34.30
10. (12hrs)(3600sec/1hr) = 43,200s
11. a. weight –down, tension- to the center, air resistance/friction – opposite of motion
b. no c. downward vector of weight
d. 2
12. e, always
13. a. vf = at à 50m/s = (10)(5s)
14. b. x =vit + 1/2at2 à 60 = 0 + ½ 3 (t2) à t =√60/1.5 = 6.32s