Answer Solution: a. 6 sec Explanation Given: s = s ( t ) = − 16 t 2 + 96 t Calculation: a. The ball strike the ground when s = s ( t ) = 0 s = s ( t ) = − 16 t 2 + 96 t = 0 t ( − 16 t + 96 ) = 0 t = 0 or − 16 t + 96 = 0 t = 0 or t = 6 discard the solution t = 0 , the strikes the ground after 6 sec . To determine To find: Instantaneous Velocity of a Ball In physics it is shown that the height s of a ball thrown straight up with an initial velocity of 96 ft/sec from ground level is, s = s ( t ) = − 16 t 2 + 96 t where t is the elapsed time that the ball is in the air. b. What is the average velocity of the ball from t = 0 to t = 2 ? Answer Solution: b. 64 ft/sec Explanation Given: s = s ( t ) = − 16 t 2 + 96 t Calculation: b. The average velocity of the ball from t = 0 to t = 2 is, Δs Δt = s ( 2 ) − s ( 0 ) 2 − 0 s ( 2 ) = − 16 × 4 + 96 × 2 = 192 – 64 = 128 s ( 0 ) = − 16 × 0 + 96 × 0 = 0 Δs Δt = s ( 2 ) − s ( 0 ) 2 − 0 = 128 − 0 2 = 64 To determine To find: Instantaneous Velocity of a Ball In physics it is shown that the height s of a ball thrown straight up with an initial velocity of 96 ft/sec from ground level is, s = s ( t ) = − 16 t 2 + 96 t where t is the elapsed time that the ball is in the air. c. What is the instantaneous velocity of the ball at time t ? Answer Solution: c. − 16 ( 2 t − 6 ) ft/sec Explanation Given: s = s ( t ) = − 16 t 2 + 96 t Calculation: c. The instantaneous velocity of the ball at time t 1 is the derivative s ’ ( t 1 ) : that is, s ’ ( t 1 ) = lim t → t 1 s ( t ) − s ( t 1 ) t − t 1 = lim t → t 1 ( − 16 t 2 − 96 t ) − ( − 16 t 1 2 − 96 t 1 ) t − t 1 = lim t → t 1 − 16 ( t 2 − t 1 2 − 6 t − 6 t 1 ) t − t 1 = lim t → t 1 − 16 [ ( t + t 1 ) ( t − t 1 ) − 6 ( t − t 1 ) ] t − t 1 = lim t → t 1 − 16 [ ( t + t 1 − 6 ) ( t − t 1 ) ] t − t 1 = lim t → t 1 − 16 ( t + t 1 − 6 ) = − 16 ( 2 t 1 – 6 ) ft/sec Replace t 1 t . the instantaneous velocity of the ball at time t is, s ’ ( t ) = − 16 ( 2 t – 6 ) ft/sec To determine To find: Instantaneous Velocity of a Ball In physics it is shown that the height s of a ball thrown straight up with an initial velocity of 96 ft/sec from ground level is, s = s ( t ) = − 16 t 2 + 96 t where t is the elapsed time that the ball is in the air. d. What is the instantaneous velocity of the ball at t = 2 ? Answer Solution: d. 32 ft/sec Explanation Given: s = s ( t ) = − 16 t 2 + 96 t Calculation: d. The instantaneous velocity of the ball at t = 2 is, s ’ ( t ) = − 16 ( 2 ( 2 ) – 6 ) = − 16 ( − 2 ) = 32 ft/sec To determine To find: Instantaneous Velocity of a Ball In physics it is shown that the height s of a ball thrown straight up with an initial velocity of 96 ft/sec from ground level is, s = s ( t ) = − 16 t 2 + 96 t where t is the elapsed time that the ball is in the air. e. When is the instantaneous velocity of the ball equal to zero? Answer Solution: e. 3 sec Explanation Given: s = s ( t ) = − 16 t 2 + 96 t Calculation: e. The instantaneous velocity of the ball is zero when, s ’ ( t ) = 0 − 16 ( 2 t – 6 ) = 0 t = 6 2 = 3 sec To determine To find: Instantaneous Velocity of a Ball In physics it is shown that the height s of a ball thrown straight up with an initial velocity of 96 ft/sec from ground level is, s = s ( t ) = − 16 t 2 + 96 t where t is the elapsed time that the ball is in the air. f. How high is the ball when its instantaneous velocity equals zero? Answer Solution: f. 144 ft Explanation Given: s = s ( t ) = − 16 t 2 + 96 t Calculation: f. How high is the ball when its instantaneous velocity equals zero. The instantaneous velocity of the ball is zero when, s ’ ( t ) = 0 − 16 ( 2 t – 6 ) = 0 t = 6 2 = 3 sec s ( 3 ) = − 16 t 2 + 96 t = − 16 ( 9 ) + 96 ( 3 ) = − 144 + 288 = 144 ft To determine To find: Instantaneous Velocity of a Ball In physics it is shown that the height s of a ball thrown straight up with an initial velocity of 96 ft/sec from ground level is, s = s ( t ) = − 16 t 2 + 96 t where t is the elapsed time that the ball is in the air. g. What is the instantaneous velocity of the ball when it strikes the ground? Answer Solution: g. − 96 ft/sec Explanation Given: s = s ( t ) = − 16 t 2 + 96 t Calculation: g. The ball strikes the ground when t = 6 , the instantaneous velocity when t = 6 is, s ’ ( 6 ) = − 16 ( 2 ( 6 ) – 6 ) = − 96 ft/sec The velocity of the ball at t = 6 sec is – 96 ft/sec , the negative value implies that the ball is travelling downward.

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ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Question

Chapter 14.4, Problem 47AYU

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight up with an initial velocity of $96 ft/sec$ from ground level is,

$s = s (t) = - 16 t^{2} + 96 t$

where $t$ is the elapsed time that the ball is in the air.

a. When does the ball strike the ground? That is, how long is the ball in the air?

Expert Solution

Answer to Problem 47AYU

Solution:

a. $6 sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} + 96 t$

Calculation:

a. The ball strike the ground when $s = s (t) = 0$

$s = s (t) = - 16 t^{2} + 96 t = 0$

$t (- 16 t + 96) = 0$

$t = 0 or - 16 t + 96 = 0$

$t = 0 or t = 6$

discard the solution $t = 0$ , the strikes the ground after $6 sec$ .

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight up with an initial velocity of $96 ft/sec$ from ground level is,

$s = s (t) = - 16 t^{2} + 96 t$

where $t$ is the elapsed time that the ball is in the air.

b. What is the average velocity of the ball from $t = 0 to t = 2$ ?

Expert Solution

Answer to Problem 47AYU

Solution:

b. $64 ft/sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} + 96 t$

Calculation:

b. The average velocity of the ball from $t = 0 to t = 2$ is,

$\frac{Δs}{Δt} = \frac{s (2) - s (0)}{2 - 0}$

$s (2) = - 16 \times 4 + 96 \times 2 = 192 - 64 = 128$

$s (0) = - 16 \times 0 + 96 \times 0 = 0$

$\frac{Δs}{Δt} = \frac{s (2) - s (0)}{2 - 0} = \frac{128 - 0}{2} = 64$

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight up with an initial velocity of $96 ft/sec$ from ground level is,

$s = s (t) = - 16 t^{2} + 96 t$

where $t$ is the elapsed time that the ball is in the air.

c. What is the instantaneous velocity of the ball at time $t$ ?

Expert Solution

Answer to Problem 47AYU

Solution:

c. $- 16 (2 t - 6) ft/sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} + 96 t$

Calculation:

c. The instantaneous velocity of the ball at time $t 1$ is the derivative $s ’ (t_{1})$ : that is,

$s ’ (t_{1}) = \lim_{t \to t_{1}} \frac{s (t) - s (t_{1})}{t - t_{1}}$

$= \lim_{t \to t_{1}} \frac{(- 16 t^{2} - 96 t) - (- 16 t_{1}^{2} - 96 t_{1})}{t - t_{1}}$

$= \lim_{t \to t_{1}} \frac{- 16 (t^{2} - t_{1}^{2} - 6 t - 6 t_{1})}{t - t_{1}}$

$= \lim_{t \to t_{1}} \frac{- 16 [(t + t_{1}) (t - t_{1}) - 6 (t - t_{1})]}{t - t_{1}}$

$= \lim_{t \to t_{1}} \frac{- 16 [(t + t_{1} - 6) (t - t_{1})]}{t - t_{1}}$

$= \lim_{t \to t_{1}} - 16 (t + t_{1} - 6)$

$= - 16 (2 t_{1} - 6) ft/sec$

Replace $t_{1} t$ . the instantaneous velocity of the ball at time $t$ is,

$s ’ (t) = - 16 (2 t - 6) ft/sec$

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight up with an initial velocity of $96 ft/sec$ from ground level is,

$s = s (t) = - 16 t^{2} + 96 t$

where $t$ is the elapsed time that the ball is in the air.

d. What is the instantaneous velocity of the ball at $t = 2$ ?

Expert Solution

Answer to Problem 47AYU

Solution:

d. $32 ft/sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} + 96 t$

Calculation:

d. The instantaneous velocity of the ball at $t = 2$ is,

$s ’ (t) = - 16 (2 (2) - 6) = - 16 (- 2) = 32 ft/sec$

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight up with an initial velocity of $96 ft/sec$ from ground level is,

$s = s (t) = - 16 t^{2} + 96 t$

where $t$ is the elapsed time that the ball is in the air.

e. When is the instantaneous velocity of the ball equal to zero?

Expert Solution

Answer to Problem 47AYU

Solution:

e. $3 sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} + 96 t$

Calculation:

e. The instantaneous velocity of the ball is zero when,

$s ’ (t) = 0$

$- 16 (2 t - 6) = 0$

$t = \frac{6}{2} = 3 sec$

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight up with an initial velocity of $96 ft/sec$ from ground level is,

$s = s (t) = - 16 t^{2} + 96 t$

where $t$ is the elapsed time that the ball is in the air.

f. How high is the ball when its instantaneous velocity equals zero?

Expert Solution

Answer to Problem 47AYU

Solution:

f. $144 ft$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} + 96 t$

Calculation:

f. How high is the ball when its instantaneous velocity equals zero.

The instantaneous velocity of the ball is zero when,

$s ’ (t) = 0$

$- 16 (2 t - 6) = 0$

$t = \frac{6}{2} = 3 sec$

$s (3) = - 16 t^{2} + 96 t$

$= - 16 (9) + 96 (3)$

$= - 144 + 288 = 144 ft$

To determine

To find: Instantaneous Velocity of a Ball In physics it is shown that the height $s$ of a ball thrown straight up with an initial velocity of $96 ft/sec$ from ground level is,

$s = s (t) = - 16 t^{2} + 96 t$

where $t$ is the elapsed time that the ball is in the air.

g. What is the instantaneous velocity of the ball when it strikes the ground?

Expert Solution

Answer to Problem 47AYU

Solution:

g. $- 96 ft/sec$

Explanation of Solution

Given:

$s = s (t) = - 16 t^{2} + 96 t$

Calculation:

g. The ball strikes the ground when $t = 6$ , the instantaneous velocity when $t = 6$ is,

$s ’ (6) = - 16 (2 (6) - 6) = - 96 ft/sec$

The velocity of the ball at $t = 6 sec$ is $- 96 ft/sec$ , the negative value implies that the ball is travelling downward.

Chapter 14.4, Problem 46AYU

Chapter 14.4, Problem 48AYU

Chapter 14 Solutions

Precalculus

Ch. 14.1 - Graph f( x )={ 3x2ifx2 3ifx=2 (pp.100-102)Ch. 14.1 - Prob. 2AYU Ch. 14.1 - Prob. 3AYU Ch. 14.1 - Prob. 4AYU Ch. 14.1 - True or False lim xc f( x )=N may be described by...Ch. 14.1 - Prob. 6AYU Ch. 14.1 - lim x2 ( 4 x 3 )Ch. 14.1 - lim x3 ( 2 x 2 +1 )Ch. 14.1 - lim x0 x+1 x 2 +1 Ch. 14.1 - Prob. 10AYU

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