In 1920 the record for a certain race was 46.6 sec. In 1950 it was 46.0 sec. let r(t)= the record in the race and t= the number of years since 1920.
(a) find a linear function that fits the data r(t)= (round to the nearest hundredth)
(b)use the function in (a)to predict the record in 2003 and in 2006.What is the predited record for 2003 and 2006 roun to the neareat tenth.
(c) In what year will the predicted record be 44.7 seconds?
To find a linear function that fits the given data, we need to determine the slope and y-intercept.
Given data:
In 1920 (t=0), the record (r(t)) was 46.6 sec.
In 1950 (t=30), the record (r(t)) was 46.0 sec.
(a) Finding the slope:
Slope (m) = (change in y) / (change in x)
= (change in r(t)) / (change in t)
Change in r(t) = 46.0 - 46.6 = -0.6 (decrease of 0.6 seconds)
Change in t = 30 - 0 = 30
Slope (m) = -0.6 / 30 = -0.02
(b) Finding the y-intercept:
We can use the point-slope form of a linear equation: y - y1 = m(x - x1)
Taking the point (x1, y1) = (0, 46.6) and the slope m = -0.02,
y - 46.6 = -0.02(x - 0)
y - 46.6 = -0.02x
Rearranging the equation to get it in slope-intercept form (y = mx + b) gives:
y = -0.02x + 46.6
Now we have a linear function that fits the data: r(t) = -0.02t + 46.6.
(c) Predicting the record for 2003 and 2006:
To find the record in 2003 (t=2003-1920=83) and 2006 (t=2006-1920=86), we substitute these values into the linear equation:
For 2003: r(t) = -0.02(83) + 46.6 = 44.04 (approximately)
For 2006: r(t) = -0.02(86) + 46.6 = 44.08 (approximately)
The predicted record for 2003 is 44.0 seconds (rounded to the nearest tenth), and for 2006 is 44.1 seconds (rounded to the nearest tenth).
(d) In what year will the predicted record be 44.7 seconds?
To find the year when the predicted record is 44.7 seconds, we set up the equation:
44.7 = -0.02t + 46.6
Solving this equation for t gives:
-0.02t = 44.7 - 46.6
-0.02t = -1.9
t = -1.9 / -0.02 = 95
Adding 95 years to 1920 gives:
1920 + 95 = 2015
The predicted record of 44.7 seconds will occur in the year 2015.