5.(A.3A) The table represents some points on the graph

The height (feet) of an object moving vertically is given by s= (-16t^2)+(208t)+(156), where t is in seconds.

OverLord2011 [107]

Answer:

The object's velocity at t = 7 is -16 ft/s

The maximum height is 832 ft and it occurs when $t=\frac{13}{2}$

Step-by-step explanation:

From the information given, the equation of motion of the object is

$s(t)= -16t^2+208t+156$

For any equation of motion s(t), the instantaneous velocity at time t is given by

$v(t)=\frac{ds}{dt}$

(a) To find the object's velocity at t = 7, you must:

$v(t)=\frac{d}{dt}(-16t^2+208t+156)\\\\v(t)=-\frac{d}{dt}\left(16t^2\right)+\frac{d}{dt}\left(208t\right)+\frac{d}{dt}\left(156\right)\\\\v(t)=-32t+208+0\\\\v(t)=-32t+208$

Next, we evaluate when t = 7

$v(7)=-32(7)+208=-224+208\\\\v(7)=-16$

The object's velocity at t = 7 is -16 ft/s

(b) To find the maximum of a function we always use the derivative of the function and we set it equal zero to find the critical points.

To find the maximum height and when it occurs, we set the velocity function equal to 0 and solve for t.

$-32t+208=0\\\\-32t+208-208=0-208\\\\-32t=-208\\\\\frac{-32t}{-32}=\frac{-208}{-32}\\\\t=\frac{13}{2}$

Next, we substitute this value into the equation of motion to find the maximum height

$s(\frac{13}{2})= -16(\frac{13}{2})^2+208(\frac{13}{2})+156\\\\s(\frac{13}{2})=-676+1352+156)=832$

The maximum height is 832 ft and it occurs when $t=\frac{13}{2}$

6 0

3 years ago

Evaluate each expression

qwelly [4]

Step-by-step explanation:

a = 2
b = 3
c = 5

$\longrightarrow \rm{{\Bigg ( \dfrac{a}{b} \Bigg )}^c} \\ \\ \longrightarrow \rm{{\Bigg ( \dfrac{2}{3} \Bigg )}^5} \\ \\ \qquad \qquad \bigstar \: \boxed{ \bf{ \Bigg ({\frac{x}{y} \Bigg )}^{m} = \frac{ {x}^{m} }{ {y}^{m}} }} \\ \\ \longrightarrow \rm{ \frac{ {(2)}^{5} }{ {(3)}^{5} } } \\ \\ \longrightarrow \bf{ \frac{ 32 }{ 243} }$