A Touch of Physics

Section 1.5
Question 68

See attachment.

I want you to do 68 parts (a), (d), (e), and (f) only.
I will then use your solution steps to do the rest.

 

68. an object is thrown upward from a height 6.5ft at velocity 72ft/s
t[1]=0
t[2]=3
(a)

s=-16t^2+v[o]*t+s[o]

given:
v[o]=72ft/s
s[o] =6.5ft
t[1]=0
t[2]=3

s=-16t^2+72*t+6.5

if t[1]=0 =>s=6.5ft
if t[1]=3 s=-16*3^2+72*3+6.5=>s=78.5ft



(d)
point t[1] is (0,6.5)
point t[2] is (3,78.5)

slope is m=(78.5-6.5)/(3-0)=72/3=24


(e)

equation of secant line through t[1] and t[2] with a slope of 24, is

s=mt+b
s=24t+b ...use point t[1] is (0,6.5)
6.5=24*0+b
b=6.5
=>
s=24t+6.5


(f) see attached

 

Beautiful. I can use this reply to do the rest when time allows. Thanks again. Read my text about working for a vacationing coworker.

 

How is 70 done when an object is dropped?

 

Let me try at least one question here. Time is limited. How about 67?

Part (a)

t_1 = 1

t_2 = 2

S = -16t^2 +64t + 6

Part (b)

See attachment. Which of the two graphs is right?



Part (c)

Let A = average rate of change.

A = [f(t_2) - f(t_1)]/(t_2 - t_1)

A = (70 - 6)/(2 - 1)

A = 64

Part (d)

When t_1 = 1, s = 6.

When t_2 = 2, s = 70.

Point A = (t_1, s_1) = (1, 6).

Point B = (t_2, s_2) = (2, 70).

Let m be slope of secant line through points A and B.

m = (70 - 6)/(2 - 1)

m = 64

Part (e)

s = mt + b

I can use either point A or B. I will use point A.

Point A = (1, 6).

6 = 64(t_1) + b

I need to find b.

6 = 64(1) + b

6 - 64 = b

-58 = b

This leads to s = 64t - 58, which is the equation of the secant line through points A and B.

If this math work is correct, can you do Part (f) for me and show the graph?

 

Can you check out my thread
Using Triangle Inequality for questions 62 and 63?