Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

774,923 users

Help with Related Rates Calculus.

0 votes
A tractor is moving at the rate of 11.8 ft/s away from a building 90.8 ft high. How fast is the distance to the top of the building increasing when the tractor is 151 ft away from the base of the building?
asked Nov 4, 2014 in CALCULUS by anonymous

1 Answer

0 votes

The Speed of the Tractor = 11.8 ft/sec.

Height of the Building = 90.8 ft.

Distance travelled by the Tractor from the base of the Building = 151 ft.

Let s be the Distance from top of the building to the tractor when x = 151 ft.

Apply Pythagoras Theorem

S² = X² + h²

S² = (151)² + (90.8)²

S² = 31045.64

S = 176.19 ft.

The Distance from top of the building to the tractor when x = 151 ft is 176.19 ft.

We know that Speed = Distance / time

Time = Distance / Speed

Time taken by Tractor = 151 / 11.8

t = 12.7966 sec.

Then the Change in the distance from top of the building to the tractor is = Distance / time

= 176.19 / 12.7699

= 13.7684 ft/sec

Therefore the Change in the distance from top of the building to the tractor is 13.7684 ft/sec.

answered Nov 4, 2014 by dozey Mentor

Related questions

asked Nov 3, 2014 in CALCULUS by anonymous
asked Nov 15, 2014 in CALCULUS by anonymous
asked Oct 25, 2014 in PRECALCULUS by anonymous
asked Feb 19, 2015 in CALCULUS by anonymous
asked Feb 14, 2015 in CALCULUS by anonymous
asked Dec 13, 2014 in CALCULUS by anonymous
...