A Touch of Physics

[nycmathguy]

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.

 

[MathLover1]

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

 

[nycmathguy]

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.

 

[nycmathguy]

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.

View attachment 112

How is 70 done when an object is dropped?

 

[nycmathguy]

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

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?

 

[nycmathguy]

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

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