Olivia B.

asked • 09/17/16

Using instantaneous velocity to find an objects position.

(b) If the object is at 290 feet and its instantaneous velocity is 6 feet per minute at 22 minutes, what is the approximate position of the object at 24 minutes?
I'm not exactly sure how to use the information to find the answer its looking for. I know two minutes have passed but would the instantaneous velocity stay the same to calculate the height? I'm not sure what to do with this question. Please help me.

Arturo O.

Olivia,

Is the object in free fall, or is it moving horizontally? We need to know that in order to answer the question.
Report

09/17/16

1 Expert Answer

By:

Arturo O. answered • 09/17/16

Tutor
5.0 (66)

Experienced Physics Teacher for Physics Tutoring

See tutors like this
See tutors like this
I assume the object is in vertical motion, in free fall. Positive direction is up.

h(t) = (1/2)at2 + v0t + h0

a = -g = -32 ft/s2

Call t = 0 the time when the object is 290 ft high and moving up at 6 ft/min (i.e. the conditions at 22 min)

Then

v0 = 6 ft/min = (6 ft)/(60 s) = 0.1 ft/s

h0 = 290 ft

Then from this moment,

h(t) = -16t2 + 0.1t + 290

Then 2 minutes later (i.e. at 24 minutes), t = (2)(60 s) = 120 s

h(120) = -16(1202) + 0.1(120) + 290

h(120) = -230098 ft
Upvote • 0 Downvote
More
Report

Steven W.

tutor
I agree.  I think this is the only way to solve the problem with the given information.  In kinematics, you need to know three of the kinematic quantities to solve for a fourth.  In this case, you want to solve, effectively, for displacement, knowing initial velocity (6 ft/min) and the time of the motion (2 min).  Without knowing a third quantity, you cannot solve it with kinematics (at least, kinematics alone).  Assuming it is in free fall allows you to assume a value for gravitational acceleration, which is the third kinematic quantity you know, which then allows you to solve.
Report

09/17/16

Arturo O.

Yes, that makes sense.  By the way, I was surprised to get such a huge number for the distance the object fell, but that is what the solution gives, with the problem as stated, and interpreted as free fall.  But I wonder if the initial upward speed was really 6 ft/s instead of 6 ft/min.  At the latter speed, the object is almost standing still!  An initial upward speed of 6 ft/s will not give such a huge vertical drop.
Report

09/18/16

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.