Review Questions and practice test questions.
Chapter 2: Linear Motion.
Average speed = distance/time
Distance from starting point (a round trip has a displacement of zero)
Velocity is speed AND direction
Acceleration is the rate of change of velocity a =∆v/t
Instantaneous speed
zero
10 m/s2
10 km/h.s
Gas pedal
Non motion can’t happen
.. g is the acceleration due to gravity at the
surface of the earth.
V = gt = 50 m/s
Neither
D = ˝ gt2 or d = 5t2
T = v/a t = 40/10 = 4 4sec up and 4 sec down. T = 8
seconds
How high will it go?
Max height = 5t2 = 5 (4)2 = 80m.
How high will it be after 6 sec?
Up for 4 seconds = 80 meters
Down for 2 seconds = 5(2)2 = 20 meters
80 meters – 20 meters = 60 meters above
ground.
What is the speed when it hits the ground?
30 m/s
How fast is it going after 2 sec?
10
m/s upwad
10m/s2
What is the velocity at that point?
0 m/s
Slope
of a distance vs. time graph is VELOCITY.
Positive slope is positive velocity (forward motion). Negative slope is negative velocity(backward
motion). Zero slope is zero velocity(at rest)
Slope of a velocity vs. time graph is Acceleration.
Calculations.
D
= vt
D
= 60 x 0.5 = 30 miles
D
= vt
D
= 600 x 2 = 1200km
D
= speed x time
D
= 50 x 4 = 200km.
Note: direction only matters if we ask about
displacement.
A
= ∆v/t
A
= 100/10
A
= 10km/h s
10
sec (5 up and 5 down)
Free
fall time is 0.2 sec
D
= 5t2
D
= 5 (0.2)2 = 0.2 meters
4.5 x 1011 is bigger.
Review.
1. Do you understand the worksheet “interpreting graphs?”
2. Do you understand all of the worksheet “free fall questions”
3. Do you understand all questions on the worksheets C-D 2.1 and 2.2?
4.
Do you understand the homework problems from the book
5. Good luck!
Graph skills
1.
A ball is thrown straight up
a. Draw the Distance vs. time graph for this (see book pg.23)
b. Draw the Velocity vs. time graph for the ball.
c. Draw the acceleration vs. time graph for the ball.
Answer in class.