Question
2: The velocity in feet per second of a falling object after t seconds is vt) = 32t. What is the
velocity of a falling object after 1, 5, and 10 seconds?
velocity of a falling object after 1, 5, and 10 seconds?
Answers
Answered by
GPT 3.5
To find the velocity of a falling object after 1, 5, and 10 seconds, we can simply plug in the values of t into the given equation v(t) = 32t.
When t = 1:
v(1) = 32(1) = 32 feet per second
When t = 5:
v(5) = 32(5) = 160 feet per second
When t = 10:
v(10) = 32(10) = 320 feet per second
Therefore, the velocity of the falling object after 1 second is 32 feet per second, after 5 seconds is 160 feet per second, and after 10 seconds is 320 feet per second.
When t = 1:
v(1) = 32(1) = 32 feet per second
When t = 5:
v(5) = 32(5) = 160 feet per second
When t = 10:
v(10) = 32(10) = 320 feet per second
Therefore, the velocity of the falling object after 1 second is 32 feet per second, after 5 seconds is 160 feet per second, and after 10 seconds is 320 feet per second.
Answered by
Damon
... assuming falling in a vacuum :)
Answered by
GPT 3.5
Yes, that assumption is correct. The given equation is for the velocity of a falling object in a vacuum, where air resistance is not a factor. In reality, the velocity of a falling object would be affected by air resistance which would cause the object to fall at a slower rate.