In 1920 the record for a certain race was 46.5 seconds. in 1980 it was 45.3 seconds let R(t) = the record in the race and t=the number of years since 1920
R(t) = 46.5 - t*0.02
what is the predicted record in 2003? 44.84
what is the predicted record in 2006?44.78
what year will the record be 44.58 seconds? 2015
help me figure this out, first I subtract 46.5 - 45.3 = 1.2 next 1980- 1920 = 60 1.2/60 = 0.02
are these correct
t = year - 1920
r = m t + 46.5
in 1980 t = 1980 - 1920 = 60
r = m (60) + 46.5 = 45.3
so
m = (45.3-46.5)/60 = -0.02
so
r = -0.02 t + 46.5 Our equation agreed
in 2003:
t = 2003 - 1920 = 83 years
r = -0.02 (83) + 46.5 = 44.84 agree
now for example in what year will it be 44 seconds?
44 = -0.02 t + 46.5
0.02 t = 46.5 - 44 = 2.5
so
t = 125
125 = year - 1920
so year = 2045 (that is just an example so you can do the real problem.)
Yes, your calculations are correct.
To calculate the change in the record time, you subtract the initial record time (46.5 seconds) from the final record time (45.3 seconds), which gives you a difference of 1.2 seconds.
To calculate the change in years, you subtract the starting year (1920) from the final year (1980), which gives you a difference of 60 years.
To find the rate of change, you divide the change in the record time (1.2 seconds) by the change in years (60 years), which gives you a rate of 0.02 seconds per year.
Using this rate of change, you can predict future record times. For example, to find the predicted record in 2003, you calculate:
R(t) = 46.5 - (2003-1920)*0.02 = 44.84 seconds
Similarly, to find the predicted record in 2006, you calculate:
R(t) = 46.5 - (2006-1920)*0.02 = 44.78 seconds
And to find the year when the record will be 44.58 seconds, you can rearrange the equation to solve for t:
44.58 = 46.5 - t*0.02
t = (46.5 - 44.58)/0.02
t ≈ 78
So the record will be approximately 44.58 seconds in the year 1920 + 78 = 1998.
Hope this helps! Let me know if you have any further questions.
Yes, your calculations are correct. To find the rate at which the record time decreases per year, you subtract the initial record time from the final record time and divide it by the number of years between the two records. In this case, the calculation is as follows:
Final record time - Initial record time = 45.3 - 46.5 = -1.2
Number of years = 1980 - 1920 = 60
Rate of change = -1.2 / 60 = -0.02
So, the rate of change is -0.02 seconds per year. This means that for every additional year after 1920, the record time decreases by 0.02 seconds.
Using the information provided, we can now predict the record times for future years:
To find the predicted record time in 2003 (t = 2003 - 1920 = 83), we substitute the value of t into the equation R(t) = 46.5 - t * 0.02:
R(83) = 46.5 - 83 * 0.02 = 44.84 seconds.
To find the predicted record time in 2006 (t = 2006 - 1920 = 86):
R(86) = 46.5 - 86 * 0.02 = 44.78 seconds.
To find the year when the record time will be 44.58 seconds, we set R(t) equal to 44.58 and solve for t:
44.58 = 46.5 - t * 0.02
t * 0.02 = 46.5 - 44.58
t * 0.02 = 1.92
t = 1.92 / 0.02 = 96
Therefore, the record time of 44.58 seconds is predicted to occur in the year 1920 + 96 = 2016.