Question

when an object is dropped, the distance it falls in t seconds, assuming that air resistance is negligible, is given by s(t) = 16t² 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 4 sec, (b) how fast the hammer is traveling 4 sec after being dropped, and (c) the hammer’s acceleration after it has been falling for 4 sec.

(a) the hammer falls 256 feet in 4 seconds. (simplify your answer.)

(b) the hammer is traveling □ ft/sec 4 seconds after being dropped. (simplify your answer.)

Explanation:

Part (b)

Step 1: Recall the velocity function

The velocity \( v(t) \) is the derivative of the position function \( s(t) \). Given \( s(t) = 16t^2 \), we find the derivative using the power rule. The power rule states that if \( f(t)=t^n \), then \( f^\prime(t)=nt^{n - 1} \). So, for \( s(t)=16t^2 \), the derivative \( v(t)=s^\prime(t)=16\times2t^{2 - 1}=32t \).

Step 2: Evaluate the velocity at \( t = 4 \)

We substitute \( t = 4 \) into the velocity function \( v(t)=32t \). So, \( v(4)=32\times4 \).

Step 3: Calculate the value

\( 32\times4 = 128 \).

Step 1: Recall the acceleration function

The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \). We already found that \( v(t)=32t \). Now, we find the derivative of \( v(t) \) with respect to \( t \). Using the power rule, the derivative of \( 32t \) (where \( n = 1 \)) is \( 32\times1t^{1-1}=32 \). The acceleration function \( a(t) \) is a constant function (since the derivative of a linear function is a constant), so it does not depend on \( t \).

Step 2: Conclusion

Since the acceleration function \( a(t) = 32 \) for all \( t \), when \( t = 4 \), the acceleration is still 32.

Answer:

128

Part (c)