DaveStrider413
contestada

An astronaut drops a rock into a crater 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?

Respuesta :

peachimin

Plug in 5 and 10 to the equation.

For the first one would be d(5)=0.8(5)^2. Multiply the equation, d(5)=4^2. Then you just multiply the exponent and the answer would be, d(5)= 10

The second one, you just do the same thing and the answer would be d(10)= 64

Answer:

12m/s

Step-by-step explanation:

The distance is given by the function:

[tex]d(t)=0.8t^2[/tex]

the distance at 5 seconds is:

[tex]d(5)=0.8(5)^2\\d(5)=0.8(25)\\d(5)=20m[/tex]

and the distance at 10 seconds:

[tex]d(10)=0.8(10)^2\\d(10)=0.8(100)\\d(10)=80m[/tex]

So, the distance covered in the interval of time between 5 and 10 seconds is:

[tex]distance=d(10)-d(5)\\distance=80m-20m\\distance=60m[/tex]

All of this is done to find the average speed with the formula:

[tex]Speed=\frac{distance}{time}[/tex]

and we substitute the distance covered wich is 60m, and the time between 5 and 10 seconds which is 5 seconds.

the average speed is:

[tex]speed=\frac{60m}{5s}\\ speed=12m/s[/tex]

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