s = = xtxt 20 m 4 s s = 5 m/s not direction dependent! a b x = 20 m time t = 4 s
DESCRIPTION
v = 3 m/s at 20 0 N of E Direction required! A B s = 20 m Time t = 4 s d= 12 m 20 oTRANSCRIPT
SPEED VS VELOCITYEQ
DEFINITION OF SPEED
• SPEED IS THE DISTANCE TRAVELED PER UNIT OF TIME (A SCALAR QUANTITY).
s = = xt
20 m 4 s
s = 5 m/s
Not direction dependent!
A
Bx = 20 m
Time t = 4 s
DEFINITION OF VELOCITY
• VELOCITY IS THE RATE OF CHANGE OF DISPLACEMENT PER UNIT OF TIME. (A VECTOR QUANTITY.)
v = 3 m/s at 200 N of E
Direction required!
A
Bs = 20 m
Time t = 4 s
d= 12 m
20o
sm
txv
412
tdv
if
if
ttdd
v
EXAMPLE 1. A RUNNER RUNS 200 M, EAST, THEN CHANGES DIRECTION AND RUNS 300 M, WEST. IF THE ENTIRE TRIP TAKES 60 S, WHAT IS THE AVERAGE SPEED AND WHAT IS THE AVERAGE VELOCITY?
Recall that average speed is a function only of total distance and total time:Total distance= 200 m + 300 m = 500 m
Avg. speed
8.33 m/sDirection does not matter!
start
x1 = 200 m
x2 = 300 m
sm
txAveSpeed
60500
EXAMPLE 1 (CONT.) NOW WE FIND THE AVERAGE VELOCITY, WHICH IS THE NET DISPLACEMENT DIVIDED BY TIME. IN THIS CASE, THE DIRECTION MATTERS.
xo = 0
t = 60 sx1= +200
mxf = -100 m
x0 = 0 m; xf =200-300= -100 m Direction of final
displacement is to the left as shown.
Average velocity:
1.67 m/s, Westv
Note: Average velocity is directed to the west.
txx
v of
smsmmv /67.1
600100
Instantaneous speed: the speed at any given instant in
time.Average speed: the average of all instantaneous
speeds; found simply by a distance/time ratio.
GUIDED AND INDIVIDUAL PRACTICE