Average Rate of Change Word Problem Help?

[madeeeeee]

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?

 

[HallsofIvy]

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.