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An astronaut drops a rock off the edge of a cliff on the Moon.
The distance, d(t), in meters, the rock travels after t seconds can be
modeled by the function d(t) = 0.8t^2. What is the average speed,
in meters per second, of the rock between 5 and 10 seconds after
it was dropped?
(1) 12 (3) 60
(2) 20 (4) 80

Respuesta :

     Using the Definition fo Average Speed, we have:

[tex]\Delta S (5)=0.8*5^2 \\ \Delta S (10)=20m \\ \\ \Delta S (10)=0.8*10^2 \\ \Delta S (10)=80m \\ \\ \Delta v= \frac{\Delta S (10) - \Delta S (5)}{\Delta t} \\ \Delta v= \frac{80-20 }{10-5} \\ \boxed {\Delta v= 12}[/tex]

Number (1)

If you notice any mistake in my english, please let me know, because i am not native.

The average speed in m/s of the rock between 5 and 10 seconds is given by;

Option 1; 12

We are given the function that represents distance in time t as;

d(t) = 0.8t²

At t = 5 seconds;

d(5) = 0.8(5²)

d(5) = 0.8 × 25

d(5) = 20 m

At t = 10 seconds;

d(10) = 0.8(10²)

d(10) = 0.8 × 100

d(10) = 80 m

Formula for average speed between two times is given as;

Average speed = change in distance/change in time

Thus;

Average speed = (80 - 20)/(10 - 5)

Average speed = 60/5

Average Speed = 12 m/s

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