If a freely falling rock were equipped with a speedometer, by how much would its speed readings increase with each second IF it were on a planet where g = 20 m/s2?
Show all work leading to your answer.
g=20 m/s^2
20 m/s
V=gt
S=g( T ^2) / 2
Is this right ?
V = g*t
V = 20*1 = 20 m/s.
V = 20*2 = 40 m/s.
V = 20*3 = 60 m/s.
Increase = 20 m/s with each second.
You are on the right track! Let's go step by step to calculate the increase in speed with each second for a freely falling rock on a planet with a gravitational acceleration of 20 m/s^2.
The acceleration due to gravity (g) can be defined as 20 m/s^2. This means that for every second that passes, the velocity (V) of the rock will increase by 20 m/s.
Using the formula V = gt, where g is the acceleration due to gravity and t is the time in seconds, we can calculate the change in velocity:
V = 20 m/s^2 * t
So, after 1 second (t = 1), the velocity would be:
V = 20 m/s^2 * 1 = 20 m/s
Similarly, after 2 seconds (t = 2), the velocity would be:
V = 20 m/s^2 * 2 = 40 m/s
As you can see, the velocity increases by 20 m/s with each passing second.
Now, let's derive the equation for the distance traveled in time (S). The equation is given by:
S = 0.5 * g * t^2
Where S is the distance and t is the time in seconds.
Using the given acceleration due to gravity (g = 20 m/s^2), we can calculate the distance traveled in time:
S = 0.5 * 20 m/s^2 * t^2
So, after 1 second (t = 1), the distance covered would be:
S = 0.5 * 20 m/s^2 * (1)^2 = 0.5 * 20 m = 10 m
Similarly, after 2 seconds (t = 2), the distance covered would be:
S = 0.5 * 20 m/s^2 * (2)^2 = 0.5 * 20 m * 4 = 40 m
Again, we can observe that the distance traveled increases with time.
In summary, the speed readings of the falling rock would increase by 20 m/s with each second passing on a planet with a gravitational acceleration of 20 m/s^2.