Question

A rock dropped on the moon will increase its speed from 0 m/s (its starting

speed) to 8.15 m/s in about 5 seconds. If you were standing on the moon,
measuring the rock's motion, what would you measure for the magnitude of
its acceleration?
O
A. 8.2 m/s2
O B. 41 m/s2
O C. 9.8 m/s2
O D. 1.63 m/s2

Answer 1

Hello!

The answer is:

The correct option is the option D

`g=1.63(m)/(s^(2))`

Why?

To calculate the acceleration, we need to use the following formula that involves the given information (initial speed, final speed, and time).

We need to use the following free fall equation:

`V_(f)=V_(o)-g*t`

Where:

`V_(f)` is the final speed.

`V_(o)` is the initial speed.

g is the acceleration due to gravity.

t is the time.

We are given the following information:

`V_(f)=8.15(m)/(s)`

`V_(o)=0(m)/(s)`

`t=5seconds`

Then, using the formula to isolate the acceleration, we have:

`V_(f)=V_(o)-g*t`

`V_(f)=V_(o)-g*t\\\\g*t=V_(o)-V_(f)\\\\g=(V_(o)-V_(f))/(t)`

Now, substituting we have:

`g=(V_(o)-V_(f))/(t)`

`g=(0-8.15(m)/(s))/(5seconds)=-1.63(m)/(s^(2))`

Therefore, since we are looking for a magnitude, we have that the obtained value will be positive, so:

`g=1.63(m)/(s^(2))`

Hence, the correct option is the option D

`g=1.63(m)/(s^(2))`

Have a nice day!