1. Describing Motion
Some More Equations…

2. d = vt d d EQUATIONS FOR MOTION UNDER CONSTANT SPEED:
CONSTANT ACCELERATION: d d

3. a) What is its acceleration?
Example 1: A car starts from rest and accelerates to a velocity of 30 m/s in 5 sec.
a) What is its acceleration?
Givens:
Base formula/rule:
Answer:
Work:
Diagrams:
Vo = 0 m/s
Vf = 30 m/s
t = 5 s
a= ?
a= 6 m/s/s
a = (30 m/s-0m/s) 5s
a= 6 m/s/s

4. d d = (0)(5s) + ½ (6m/s/s)(5s)2 d = 0 + 75 m
Example 1: A car starts from rest and accelerates to a velocity of 30 m/s in 5 sec.
B) How far did the car travel during this 5 sec?
Givens:
Base formula/rule:
Answer:
Work:
Diagrams:
Vo = 0 m/s
Vf = 30 m/s
t = 5 s
a = 6 m/s/s
d = ? d
d= 75 m d
d = (0)(5s) + ½ (6m/s/s)(5s)2
d = m

5. 2d a = (0 m/s)2 - (20 m/s)2 2(70m) a = - 2.85714 d
Example 2: A train traveling at 20 m/s comes to a stop at the station in 70 m. What is the deceleration rate of the train?
Givens:
Base formula/rule:
Answer:
Work:
Diagrams:
Vo = 20 m/s
Vf = 0 m/s
d = 70 m
a = ? d
A = m/s/s
a = Vf 2 –Vo 2 2d
a = (0 m/s)2 - (20 m/s)2
2(70m)
a =

6. A car traveling at velocity v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is initially traveling at 2v ? Assume that the acceleration due to the braking is the same in both cases.
Solve using reasoning, not necessarily formulas

7. The car will need an amount of distance equal to 4d to stop if the initial velocity is 2v. Since the acceleration is the same in both cases and the final velocity is zero in both cases, the stopping distance is determined by the initial velocity. The equation vf2 = vi2 + 2ad indicates that the initial speed squared is directly proportional to the stopping distance. Since vi was 2v, the square of 2v is 4v2 so stopping distance is four times as great as the initial stopping distance.

8. Describing Motion Class/Homework: This Way & That Way II
Motion Problems #7-11