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Question 361497: Trying to help my daughter with Question #54 on page 299 of McDougal Little Algebra 1.
An In-line skater practices ar a race track. In two trials the skater travles the same distance going from a standstill to his top racing speed. He then travels at his top racing speed for different distances.
Trial 1, Time at Top Speed 24 seconds, total distance traveled 300 meters.
Trial 2, Time at Top Speed 29 seconds, Total distance traveled 350 meters.
A. Model -Write an equation that gives the total distance traveled (in meters) as a function of time (in seconds) at top racing speed.
B. - Justify -What do the rate of change and initial value in your equation represent? Explain you answer using unit analysis.
C. Predict - One lap around the race track is 200 meters. The skater starts at a standstill and completes 3 laps. Predict the number of seconds the skater travels at his top racing speed. Explain your method.
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! the extra 5 seconds at top speed (29-24) resulted in 50 more meters of travel (350-300)
___ so the top speed is ___ 50m / 5s or 10 m/s
10 m/s for 24 s is a distance of 240 m
___ this means that it takes 60 m (300-240) to get to top speed
A. d = 10t + 60
B. the rate of change represents the top speed , and the initial value represents the speed-up distance
___ meters = (meters/sec)(sec) + meters
C. 3 laps is 600m (3 * 200)
___ 60 m is used for speed-up , so 540m is the distance at top speed
___ 540m at 10 m/s means 54 sec at top speed
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