Question

A rock dropped on the moon will increase its speed from 0 m/s ( its starting speed when first dropped) to 4.9 m/s in about 3 seconds. What value do you calculate for the magnitude of the acceleration of gravity on the moon?

Answer 1

Acceleration = (change in speed) divided by (time for the change).

From the figures given in the question, the acceleration is ( 49/3 ) =

16.33 m/sec2

.

There's no way that this is happening on the moon. That acceleration is about 67%

greater

than the acceleration of gravity on the earth's surface. It should be about 83%

less,

or about

1.63 m/sec

2

.

I see the problem now. The '49' in the question should be '

4.9

'.

apex- 1.63 m/s2

Answer 2

Acceleration, a = 1.64 m/s²

Step-by-step explanation:

It is given that,

A rock is dropped on the moon, its initial velocity u = 0 m/s

Its speed increases from 0 to 4.9 m/s, v = 4.9 m/s

Time taken, t = 3 seconds

We have to find the magnitude of the acceleration due to gravity on the moon. The magnitude of acceleration is given by :

Acceleration, a = rate of change of velocity

`a=(v-u)/(t)`

`a=(4.9\ m/s-0)/(3\ s)`

a = 1.64 m/s²

So, the value of acceleration of gravity on the moon is 1.64 m/s². Hence, this is the required solution.