Question

this is section 2.4 problem 24:
if a rock is thrown from the ground with an initial velocity of 120 ft/sec, then the height of the rock, in feet, at t seconds can be modeled by s(t)=120t−16t², 0≤t≤7.5 .
(a) the velocity is v(t) = .
(b) v(3) = ---select--- . it represents the ---select--- of the rock at 3 seconds after it is thrown.
(c) at t=3 seconds, the rock is ---select--- since its velocity is ---select--- .
hint: follow example 6.
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ebook

Explanation:

Part (a)

Step1: Recall velocity as derivative of position

The velocity function \( v(t) \) is the derivative of the position function \( s(t) \). Given \( s(t) = 120t - 16t^2 \), we use the power rule for differentiation: if \( f(t) = at^n \), then \( f'(t) = nat^{n - 1} \).

Step2: Differentiate \( s(t) \)

For the first term \( 120t \), the derivative is \( 120 \times 1 \times t^{1 - 1} = 120 \). For the second term \( -16t^2 \), the derivative is \( -16 \times 2 \times t^{2 - 1} = -32t \). So, \( v(t) = s'(t) = 120 - 32t \).

Step1: Substitute \( t = 3 \) into \( v(t) \)

We have \( v(t) = 120 - 32t \). Substitute \( t = 3 \): \( v(3) = 120 - 32(3) \).

Step2: Calculate the value

First, calculate \( 32(3) = 96 \). Then, \( 120 - 96 = 24 \). The velocity function \( v(t) \) represents the rate of change of the position (height) with respect to time, so \( v(3) \) is the velocity (speed with direction) of the rock at \( t = 3 \) seconds.

Step1: Analyze the sign of \( v(3) \)

We found \( v(3) = 24 \), which is positive. A positive velocity (in the context of an object thrown upward, positive velocity is upward direction) means the object is still moving upward (since velocity is positive, the height is increasing at \( t = 3 \) seconds).

Step2: Determine the motion

Since the velocity at \( t = 3 \) is positive, the rock is moving upward (or rising) because a positive velocity (in the upward - defined direction) indicates that the height is increasing with time at that instant.

Answer:

\( 120 - 32t \)

Part (b)