Average Rate of Change Word Problem Help?

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SUMMARY

The discussion focuses on calculating the average and instantaneous rates of change for a ball thrown vertically from a height of 20 meters, described by the equation h(t) = -4.9t² + 3t + 20. The average rate of change between 1 and 4 seconds is calculated as -3.6 m/s, which was initially questioned but confirmed incorrect due to miscalculation of h(4) and h(1). The instantaneous rate of change at t = 1 second is determined to be -6.8 m/s, which is validated as correct.

PREREQUISITES
  • Understanding of quadratic functions and their properties
  • Knowledge of derivatives and their application in calculating rates of change
  • Familiarity with the concept of average rate of change
  • Basic algebraic manipulation skills
NEXT STEPS
  • Review the calculation of h(4) and h(1) to confirm average rate of change
  • Study the concept of derivatives in calculus, specifically for polynomial functions
  • Practice additional problems involving average and instantaneous rates of change
  • Explore applications of quadratic equations in real-world scenarios
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Students studying calculus, educators teaching rates of change, and anyone interested in understanding motion described by quadratic equations.

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Average Rate of Change Word Problem Help!?

12. A ball is thrown vertically upward from a roof of a school that is 20 m high. The equation which gives its height above ground, in metres, after t seconds, is h(t) = - 4.9t2 + 3t + 20,
a) Calculate the average rate of change of the ball between 1 and 4 seconds. [2 marks]

answer:
h(4)-h(1) / 3

-3.6 m/s

is this right?


b) Determine the instantaneous rate of change when t = 1 s. [4 marks]

answer:
h(t) = - 4.9t2 + 3t + 20
h'(t)= -9.8t + 3
= -9.8(1) + 3
= -6.8 m/s
At t= 1s, the skydiver is falling at a rate of -6.8 m/s.

is this right?
 
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madeeeeee said:
12. A ball is thrown vertically upward from a roof of a school that is 20 m high. The equation which gives its height above ground, in metres, after t seconds, is h(t) = - 4.9t2 + 3t + 20,
a) Calculate the average rate of change of the ball between 1 and 4 seconds. [2 marks]

answer:
h(4)-h(1) / 3

-3.6 m/s

is this right?
No, it is not. What did you get for h(4) and h(1)?




b) Determine the instantaneous rate of change when t = 1 s. [4 marks]

answer:
h(t) = - 4.9t2 + 3t + 20
h'(t)= -9.8t + 3
= -9.8(1) + 3
= -6.8 m/s
At t= 1s, the skydiver is falling at a rate of -6.8 m/s.

is this right?
Yes, that is correct.
 

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