Thread: speed velocity

A person is traveling 650 fpm it takes them 25 seconds to stop how many feet did it take?

Originally Posted by dmark

A person is traveling 650 fpm it takes them 25 seconds to stop how many feet did it take?

First convert your units.

650 fpm = 650 ft/min = 650 ft/(60 s) = 10.83333 ft/s

d = vt

d = (10.83333 ft/s)(25 s) = 270.8333333 ft

Or 271 ft.

-Dan

Thank you for your answer. In this problem it is not a constant volocity but a deceling. Do I need to use the slope? y=ax+b?

Originally Posted by dmark

Thank you for your answer. In this problem it is not a constant volocity but a deceling. Do I need to use the slope? y=ax+b?

I hate it when I do that! My apologies!

Originally Posted by dmark

A person is traveling 650 fpm it takes them 25 seconds to stop how many feet did it take?

Let's try this again:

We don't know the acceleration so let's use an equation that doesn't involve the acceleration:
x = x0 + (1/2)(v0 + v)t

I'm envisioning a coordinate system where + x is in the direction of motion and the origin is at where the person starts to decelerate. Thus x0 = 0 ft, v0 = 650 ft/min, v = 0 ft/min. Using the unit conversion from my first post we get that
v0 = 10.833333 ft/s.

Thus:
x = (1/2)(v0)t = (1/2)(10.833333 ft/s)(25 s) = 135.417 ft

So 135 ft.

-Dan