On the Moon, a falling object falls just 2.65 feet in the first second after being dropped. Each second it falls 5.3 feet farther than in the previous second. How far would an object fall in the first ten seconds after being dropped?
S10 = 10/2 (2*2.65 + 9*5.3) = 265.0
To find the total distance the object would fall in the first ten seconds, you can calculate the sum of an arithmetic sequence.
The first term (a₁) is given as 2.65 feet, and the common difference (d) is 5.3 feet, since the object falls 5.3 feet farther each second.
The formula for the sum of an arithmetic sequence is:
Sₙ = (n/2) * (2a₁ + (n-1)d)
Where Sₙ represents the sum of the first n terms.
Plugging in the given values:
S₁₀ = (10/2) * (2 * 2.65 + (10-1) * 5.3)
Simplifying the expression:
S₁₀ = 5 * (5.3 + 9 * 5.3)
S₁₀ = 5 * (5.3 + 47.7)
S₁₀ = 5 * 53
S₁₀ = 265
Therefore, an object would fall 265 feet in the first ten seconds after being dropped on the Moon.
To find out how far an object would fall in the first ten seconds on the Moon, we can use a mathematical formula for the distance traveled by a falling object. The formula is:
d = ut + (1/2)gt^2
where:
d = distance traveled
u = initial velocity (which is 0 in this case since the object is dropped)
t = time in seconds
g = acceleration due to gravity (which is approximately 1.62 ft/s^2 on the Moon)
Let's break down the problem step by step:
Step 1: Find the initial distance traveled
We are given that the object falls 2.65 feet in the first second. Since it is dropped, the initial velocity (u) is 0. Therefore, we can say that the distance traveled in the first second (t = 1) is 2.65 feet.
Step 2: Find the incremental increase in distance each second
We are told that each second the object falls 5.3 feet farther than in the previous second. This means that for every subsequent second, the object will fall (2.65 + 5.3) feet more than the previous second.
Step 3: Calculate the distance for each second
Using the formula d = ut + (1/2)gt^2, we can calculate the distance traveled in each second and add it up for the first ten seconds.
Let's calculate:
- For the first second (t = 1):
d1 = 0(1) + (1/2)(1.62)(1^2) = 0 + 0.81 = 0.81 feet
- For the second second (t = 2):
d2 = 0(2) + (1/2)(1.62)(2^2) = 0 + 1.62(4) = 0 + 6.48 = 6.48 feet
- For the third second (t = 3):
d3 = 0(3) + (1/2)(1.62)(3^2) = 0 + 1.62(9) = 0 + 14.58 = 14.58 feet
We can continue this process for the remaining seconds up to ten.
Step 4: Calculate the total distance for the first ten seconds
To find the total distance traveled in the first ten seconds, we add up the distances calculated for each second:
Total distance = d1 + d2 + d3 + ... + d10
Calculate the remaining distances using the formula d = ut + (1/2)gt^2 and add them up to find the total distance traveled in the first ten seconds.
By following these steps and performing the necessary calculations, you can find the total distance an object would fall in the first ten seconds after being dropped on the Moon.