anyone good with acceleration and junk like that?

Question

A small fish is dropped by a pelican that is rising steadily at 0.50 m/s. After 5.0 s, the velocity of the fish is -48.5 m/s

How far below the pelican is the fish after 5.0 s?

how do you feel sitting at your computer answering random people's questions all dya huh? dont you feel like you need to get out of the house?

im a junior in high school

im a junior in high school

Answer 1

wat grade u in?

The information provided makes absolutely no sense

Answer 2

You actually have redundant information here. The velocity of the fish vf = .5-at, where a=acceleration. For t=5, vf=-48.5. Thus we can solve for a -48.5 = .5 - a*5 5a=49 a=49/5 = 9.8m/sec^2. We should already know this, because that is g, the acceleration of a falling body on earth.

The distance the fish falls in 5 seconds D = integral(0 to 5) of [vf*dt]. substituting for vf, we get D = integral0 to 5) of [.5 - 9.8*t] = .5*t - .5*9.8*t^2 or 2.5 - .5*9.8*25 = 2.5 - 122.5 = -120m. The negative sign means the downward direction, since the pelican velocity upward is positive. Meanwhile the pelican has risen 5*.5 = 2.5m, so the total distance between them is 2.5 - (-120) = 122.5m

EDIT: I just read that you are a junior in high school so you may not be familiar with integrals yet. In which case you have to know the velocty/distance/acceleration/time formulas. For constant velocity, s=vt; for constant acceleration s=.5a^t^2. Velocities add algebraically, meaning an up velocity (+) minus a down velocity (-) is the arthmetic sum of the two.

Answer 3

Pelican rising at .5m/s over 5 seconds. (0.5*5)=2.5 meters

velocity (force of gravity on falling body)= -9.8 meters per second^2

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