Question

When an object is dropped on a certain earth-like planet; the distance it falls in t seconds, assuming that air resistance is negligible, is given by s(t) = 19t 2 where​ s(t) is in feet. Suppose that a​ medic's reflex hammer is dropped from a hovering helicopter. Find ​(a) how far the hammer falls in 3 ​sec, ​(b) how fast the hammer is traveling 3 sec after being​ dropped, and ​(c) the​ hammer's acceleration after it has been falling for 3 sec.

Answer 1

The hammer falls 171 feet in 3 seconds, travels at a speed of 114 feet per second after 3 seconds, and has a constant acceleration of 38 feet/second^2.

Step-by-step explanation:

We are given the distance-time function for an object falling on an earth-like planet without air resistance: s(t) = 19t2. Let's calculate the required values.

(a) Distance fallen in 3 seconds

Plugging in t = 3 seconds into the given function:

s(3) = 19(3)2 = 19(9) = 171 feet.

(b) Velocity after 3 seconds

The velocity function is the first derivative of the distance-time function, v(t) = s'(t). So:

v(t) = 2(19)t = 38t.

v(3) = 38(3) = 114 feet/second.

(c) Acceleration after 3 seconds

Acceleration is the second derivative of the distance-time function or the first derivative of the velocity function, which is constant in uniform acceleration:

a(t) = v'(t) = 38 feet/second2.

The acceleration is 38 feet/second2, which is constant and independent of the time.

Answer 2

171 ft

114 ft/s

38 ft/s²

Step-by-step explanation:

The function is
`s(t)=19t^2`

At s = 3 s

`s(3)=19* 3^2\\\Rightarrow s(3)=171\ ft`

The hammer falls 171 ft

Differentiating the function with respect to time we have

`v=(d)/(dt)19t^2\\\Rightarrow v=38t`

at t = 3 s

`v=38* 3\\\Rightarrow v=114\ ft/s`

The velocity of the hammer is 114 ft/s

Differentiating with v respect to t

`a=(d)/(dt)38t\\\Rightarrow a=38\ ft/s^2`

The acceleration is 38 ft/s²