A helicopter is rising straight up in the air. The distance from the ground t se

Question

A helicopter is rising straight up in the air. The distance from the ground t seconds after takeoff is s(t) feet, where s(t) = t^2+t

A) How high did the helicopter rise during the 3rd second?

B) How high did the helicopter rise during the 3rd second?

C)What is the average velocity of the helicopter during the first 3 seconds

D) what is the velocity of the helicopter after 3 seconds

E) what is the acceleration after 3 seconds?

F) how fast is the helicopter rising if its 20 feet of the ground

G) how far off the ground is the helicopter if its risisng velocity at a rate of 13feet/second?

Explanation / Answer

Ok, for part a, substitute 20 = s. So:

20 = t^2 + t
0 = t^2 + t - 20

Using the quadratic formula, that gives t = 4 and t = -5 (obviously second answer is invalid). So part a is 4 seconds.

b) Since you are given the distance formula, you can differentiate to give the velocity (1st deriv.) and acceleration formula (2nd deriv.). Hence:

v(t) = 2t + 1
Subbing in t = 4 (as found in part a):
v(4) = 2 x 4 + 1
v(4) = 9 ft/s

Differentiate again to get acceleration formula:
a(t) = 2 ft/s/s

Therefore, the helicopter has an acceleration of 2, regardless of time!

Hope that helps!

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