Question

A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t2 + 8 underroot t, where t is the time in seconds and s(t) is the distance in feet from the top of the building. A) Write a formula for the average velocity of the ball for t near 4. B) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds. C) What is your estimate for the instantaneous velocity of the ball at t = 4.

Answer 1

Answer:

Explanation:

s(t) = -16t2 + 8√t

A )

Average velocity

s(t) / t = (-16t2 + 8√t)/t

A(t)= -16t +  8 / √t

average velocity of the ball for t near 4.

A(t) =  -16t +  8 / √t

          Lt t⇒4

B )

Distance covered in 4 s

-16t2 + 8√t

= - 16 x 16 + 8 x 2

= - 240

Distance covered in 5 s

= - 16 x 25 + 8 √5

= -400 + 17.88

= -382.12

distance covered in duration from 4 to 5 sec

= -142.12

velocity = - 142.12 / 1 = - 142.12 m /s

Distance covered in 4.5 s

= -16 x 4.5² + 8√4.5

= -324 + 16.97

= -307

distance covered during 4 to 4.5

= 67

velocity during 4 to 4.5

= -67 / .5

- 134 m /s

distance covered in 4.05 s

-16 x 4.05² + 8√4.05

-262.44 + 16.1

-246.34

distance covered during 4 to 4.05

= -6.34

velocity during 4 to 4.05

= -6.34 / .05

- 126.8 m /s

C )

Instantaneous velocity at t = 4

= - 120 m /s

Answer 2

(A) As 't' approaches to 4s, the formula of average velocity is

$v_{avg} _{|(t\rightarrow4)}=Lim_{|(t\rightarrow4)}[-16t+ \frac{8}{\sqrt{t}} ]$

(B) The average velocity for the time interval starting at t = 4 with a duration of 1 second is $142.11\,m/s$.

The average velocity for the time interval starting at t = 4 with a duration of 0.5 seconds is $-134.058\,m/s$

The average velocity for the time interval starting at t = 4 with a duration of 0.05 seconds is $-126.8\,m/s$.

(C) The instantaneous velocity of the ball at 4s is $-126\,m/s$.

The answers are explained as follows.

Given that the distance is a function of time.

$s(t)=-16\,t^2\,+\,8\sqrt{t}$

(A) The average velocity can be given by,

$v_{avg}=\frac{-16\,t^2\,+\,8\sqrt{t} }{t} =-16t+ \frac{8}{\sqrt{t}}$

As 't' approaches to 4s, the formula becomes;

$v_{avg} _{|(t\rightarrow4)}=Lim_{|(t\rightarrow4)}[-16t+ \frac{8}{\sqrt{t}} ]$

(B) We know that the average velocity or in this case speed is the total distance by the total time taken.

Distance covered in 4 s can be found by putting $t=4s$ in the distance formula.

• $s(4)=(-16\times16)+(8\times2)=-240\,m$

Distance covered in 5 s can also be found by the same method

• $s(5)=(-16\times25)+(8\times\sqrt{5} )=-382.11\,m$

Therefore, the distance covered in from 4 to 5 seconds is;

• $s(5) -s(4)=-382.11\,m-(-240\,m)=-142.11\,m$

• So, the average velocity here = $\frac{-142.11\,m}{5\,s-4\,s}=-142.11\,m/s$

Distance covered in 4.5 s is given by,

• $s(4.5)=(-16\times4.5^2)+(8\times\sqrt{4.5} )=-307.029\,m$

Therefore, the distance covered in 4 to 4.5 seconds is;

• $s(4.5)-s(4)=-307.029\,m-(-240\,m)=67.029\,m$

• So, the average velocity here = $\frac{-67.029\,m}{4.5\,s-4\,s}=-134.058\,m/s$

Distance covered in 4.05 s is given by,

• $s(4.05)=(-16\times4.05^2)+(8\times\sqrt{4.05} )=-246.34\,m$

Therefore, the distance covered in 4 to 4.05 seconds is;

• $s(4.05)-s(4)=-246.34\,m-(-240\,m)=-6.34\,m$

• So, the average velocity here = $\frac{-6.34\,m}{4.05\,s-4\,s}=-126.8\,m/s$

(C) The instantaneous velocity of the ball can be found by differentiating the function $s(t)$.

• $v(t)=\frac{ds(t)}{dt} =\frac{d}{dt}(-16\,t^2\,+\,8\sqrt{t})=-32t+\frac{4}{\sqrt{t} }$
• For the instantaneous velocity of the ball at 4s, substitute $t=4\,s$ in the above equation.

• $v(t)=(-32\times4)+\frac{4}{2}=-126\,m/s$

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