Question

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 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?

Answer 1

Answer: $12\text{ metre per seconds}$

Step-by-step explanation:

Given equation: $d(t)=0.8t^2$ , d(t), in meters, the rock travels after t seconds .

At t=5

The rock travels distance=$d(5)=0.8(5)^2=0.8\times25=20\ m$

At t=10

The rock travels distance=$d(10)=0.8(10)^2=0.8\times100=80\ m$

Total distance traveled by rock 5 and 10 seconds after it was dropped

=80-20=60 m

Total time =10-5=5 seconds

We know that the average speed =$\frac{\text{Total distance}}{\text{Total time}}$

$=\frac{60}{5}=12\text{ meters per second}$