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?

Answer 1

... assuming falling in a vacuum :)

Answer 2

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.

Answer 3

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.