Unformatted text preview: 30 25 m/sec 25
S
P
E
E
D E D 20
15
10
5
A
(0 , 0 ) 5 B
10
15
TIME (sec) C
20 25 t DISTANCE = SPEED * TIME [meters = meters * seconds]
seconds
over the interval [ t = 10 secs, t = 20 secs] the SPEED = 25 m/sec ( constant ). Distance travelled from t = 10 secs to t = 20 secs is 25m/sec ∗ 10 sec = 250m
which is the ar ea under the speed cur v e = area of rectangle BCDE.
area
curv
ar
In the case of distance travelled between t = 0 secs and t = 10 secs, the
travelled
SPEED is NOT constant. What SPEED are you going to use?
181 By looking at the graph we know the SPEED at each and every INSTANT
between t = 0 secs and t = 10 secs. So we have to take the product of
SPEED and TIME at each and ever y INSTANT and sum it up. This in effect
will give the ar ea under the speed cur v e from t = 0 secs to t = 10 secs
area
ar
which is the area of triangle ABE.
This is 1 base * height = 1 * 10 secs * 25 m / s ec = 125 meters .
2
2
Conceptually we note that from the speed cur v e ( INSTANTANEOUS RATE
speed
OF CHANGE of position graph) we got the CHANGE in position. CHANGE in position
position
has both sign and magnitude. In this case we have denoted going East (right) as
positiv
neg tiv
positive and going West (left) as negative on the xaxis. We may denote the SPEED
ne
of the car by x’(t) and hence its POSITION on the xaxis will be x(t).
Because the acceleration is steady or constant over [0, 10) we may use the average
acceleration over [0, 10).
x’(10) − x’(0) 2.5m
/sec2
acceleration over [0, 10) : x”(t) =
=
10 secs
acceleration over t ≥ 10 : x”(t) = 0
2.5t2/ over [0, 10]
speed over [0, 10] :
2
x(t) =
2.5t m/
25t − 125 for t ≥ 10
x’(t) = ∫x”(t)dt =
sec
speed over t ≥ 10 :
375
x’(t) = 25 m/sec
250
position over [0, 10] :
2
125
x(t) = ∫ 2.5t dt = 2.5t / 2
position over t ≥ 10 :
[meters] { x(t) = ∫ 25 dt = 25t − 125 0 10 20 [secs] t From the nature of the problem at t = 10 seconds we may determine that the
CONSTANT OF INTEGRATION is −125 meters...
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 Fall '09
 TAMERDOğAN
 Limit, Δx

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