Wed 16 SEP:
Key concept: Motion Equations & graphs
Quiz (1st Equations) : (Monday 20 Sep)
1. (Relationship between work and energy)
W =
2. (net force – Newton’s 2nd law)
SF =
3. (Definition of acceleration- relationship with velocity)
a =
4. (Distance after a period of time)
Xf =
5. (Velocity after a period of time)
Vf=
6. (Velocity/distance/acceleration without time)
Vf2 =
7. (Mechanical Work)
W =
8. (Force of friction)
Ffrict =
9. (Kinetic Energy)
KE =
10. (Potential Energy)
DUg =
11. (Definition of Linear momentum)
r =
12. (Definition of change of momentum and Impulse)
Dr =
Journal: You push on a box causing an acceleration. Newton's 3 law says that the box pushes back on you with the same force that you push with.
a. Does that mean that no work is accomplished?
b. Explain the difference between Newton's 2nd and 3rd laws in this event.
Put in Scientific Notation:
A
1. .35 = 3.5 x 10 -1
2. 13,500 = 1.35 x 10 4
3. .0006200 = 6.200 x 10 -4
4. 2,400 x 102 = 2.4 x 10 5
5. 4,030,000 x 10-6 = 4.03 x 100
6. Solve to one significant figure (you don’t need a calculator):
(37,153 x 10 20 )(0.0096 x 10 -12) =
(167.6 x 10 -32)
2 x 10 40
7. 156.6 km/hr to m/s = 43.5 m/s
8. 47 m3 to cm3 = 47,000,000cm3
9. 52 ∟1500
X = - 45, y = 26
10. 40 ∟323.10
X = 32 , y = - 24
11. Put into polar form:
x1 = 10, y1 = -7
x2 = -12, y2 = 23
x3 = -6, y3 = 2
x = -8 y = 18 à
à 19.7 ∟1140
12. Put into polar form (show diagram):
X = 90, y = 150 175∟59 0
13. You bike east @ 20km/hr due for 30mins. Then head south @ 30km/hr for 1.0 hr. Then you head east again @ 20km/hr for 30mins. Finally, you head west at 40 km/hr for 2 hrs.
a. Speed 32.5 km/hr
b. Displacement
67.1 ∟206.6 0
c.Velocity
16.8 ∟206.6 0
Problems:
1. a. How far did the cart travel in the first 10 minutes?
b. What was its average acceleration during this time interval?
2. What was its average acceleration between 10 and 15 minutes?
3. Briefly describe its behavior between 10 and 15 minutes?
1. a. displacement = 1/2bh = 1/2(10)(60) = 300m
b. a = DV/t = 60/10 = 6m/min2
2. a = DV/t = 0/5 = 0m/min2
3.constant velocity, no acceleration, displacement is increasing & positive
4. How far did it travel between 10 and 30 minutes?
5. a. How far did it travel between 30 and 40 minutes?
b. What was its average acceleration between 30 and 40 minutes?
6. How far did it travel between 40 and 55 minutes?
7. What was its final displacement?
8. a. What was the cart's average speed 0 to 55 mins?
b. What was its average acceleration 0 to 55 mins?
9. What would a position vs time graph look like between 15 to 30 seconds
4. displacement =
10-15 min: bh = (5)(60) = 300 m
15-30 min: 1/2bh = 1/2(15)(60) = 450 m
5. a. displacement =1/2bh = 1/2(10)(-40) = -200 m
b. a = DV/t = -40/10 = -4m/s2
6. displacement =1/2bh = 1/2(15)(-40) = -300 m
7. xf = total area under curve
xf = 300m + 300m + 450m + (-200) + (-300)
xf = 550m
8. a. speed = total distance/t = 1550m/55min
speed = 28.2 m/s
b. a = DV/t = 0/55 = 0m/s2
9.
More Problems
1. A hawk can dive @ 70m/s. How long to reach the ground from a height of 200M?
x = (v)(t) à 200 = 70 t à t = 2.86s
2. A torpedo is fired with a constant speed of 30 m/s at a target. After traveling 900m at that speed, the torpedo runs out of fuel and has a negative acceleration of -0.5m/s2.
Find:
a. time to stop (from firing pt)
b. Total distance to stop from firing pt.
2. (time at constant V)
x = vt à 900 = 30t à t = 30s
(time at neg acceleration)
Vf = Vo + at à 0 = 30 + -.5t à t = 60s
a. Total time:
30s + 60s = 90s
b. (total dist) = constant v (900m)+ neg acc
(-acc x) X = Vot + ½ at2 à
= (30x60) +.5(-0.5)(602)
= 1800 + (-900) à
x = 900m
Total dist = 900m + 900m = 1800m
3. A girl throws a rock horizontally, with a velocity of 25 m/s, from a bridge. It falls 100 m to the water below. How far does the rock travel horizontally before striking the water?
3. No acceleration in the horizontal (x) direction, only in the y direction:
X distance depends on two things: a.) time b.) horizontal speed.
Time: a.) Xy = Vot + ½ at2 à (-100) = (0) + .5 (-10)(t2) à t = 4.5sec
b.) Xx = Vot + ½ at2 à Xx = (25m/s) (4.5s) = 112 m
4. An object is thrown upward with a speed of 20 m/s on the surface of planet B where the acceleration of gravity is 4.0 m/s2. How long does it take for the object to reach maximum height?
Vf = Vo + at
0 = 20 + (-4)(t) à t = 5.0s
5. A ballast bag of sand dropped from a hot-air balloon hits the ground at a certain speed and the craft slowly rises and soon comes to a stop. If a second identical bag is then dropped and it hits the ground three times as fast as the first.
How high was the balloon when it was dropped in comparison to when the first bag fell?
Vf2 – Vo2 = 2aDX (if the V is tripled and it has an exponent of 2 (Vf2 ), then dist (x) goes up 9 times.
Vf2 = x then, (3Vf)2 = 9x
7. A race car accelerates from rest covering a distance of 7.5m in the first sec (Vf = 15.0 m/s), 10m during the 2nd second (Vf = 5.0 m/s), and 50m (Vf = 95.0 m/s), during the 3rd second. Find:
a. What is the average speed and acceleration during the 1st second?
b. What is the average speed and acceleration during the 2nd second?
c. What is the average speed and acceleration during the 3rd second?
d. What is the average speed and acceleration over the first 3 seconds?
4. a) v= x0-1 /t à V = 7.5/1 à speed = 7.5m/s
a0-1 = Dv/t à 15 – 0 /1 = 15m/s2
b) v= x1-2 /t à V = 10.0/1 à speed = 10.0m/s
a1-2 = Dv/t à 5.0 – 15.0 /1 = - 10.0m/s2
c) v= x2-3 /t à V = 50.0/1 à speed = 50.0m/s
a2-3 = Dv/t à 95.0 – (5.0) /1 = 90.0m/s2
d) v= x0-3 /t à V = 67.5/3 à speed = 22.5m/s
a0-3 = Dv/t à 95.0 – 0.0 /3 = 31.7m/s2