Physics, 31.07.2019 13:30 maudiejane

Aparticle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.02t3 (a) find the velocity at time t (in ft/s). v(t) = .04t3−.06t2 (b) what is the velocity after 1 second(s)? v(1) = -.02 ft/s (c) when is the particle at rest? t = s (smaller value) t = s (larger value) (d) when is the particle moving in the positive direction? (enter your answer using interval notation.) (e) find the total distance traveled during the first 12 seconds. (round your answer to two decimal places.) ft (f) find the acceleration at time t (in ft/s2). a(t) = find the acceleration after 1 second(s). a(1) = ft/s2

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Since you solved a and b I will start with part c.
Part C
To answer this question we need to find zeros of a velocity function:
v(t)=0.04t^3-0.06t^2

We can factor this polynomial:
v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)

Now it's pretty easy to find zeros. This function will be equal to zero when any of the factors are equal zero.
$t^2=0;\\ 0.04t-0.06=0$
We solve these two equations and we get our zeros:
$t_1=0; t_2=\frac{3}{2}$
The particle is at rest at t=0 and t=3/2.
Part D
To solve this we need to determine when our velocity function is greater than zero. We will use factored form.
We determine when each factor is greater than zero and with that information, we build the following table:
$\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}$
We can see, from the table, that our function is positive when $- \infty < t$ and t>3/2.
That is the range in which particle is moving in positive direction.
Part E
We know that distance traveled is given with:
s(t)=0.01t^4 - 0.02t^3

We simply plug in t=12 to find total distance traveled:
s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft

Part F
We know that acceleration is defined as a rate of change of velocity.
We find acceleration by taking the first derivative of velocity with respect to time.
$a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t$
To find acceleration after 1 second we simply plug in t=1s in above equation:
a(1)=0.12-0.12=0

Answer from: dymi6n

The particle's acceleration at 1st sec is 0 while after 12 seconds particle move 172.8 ft.

Given here,

The distance traveled,

$\rm \bold{S_(_t_) = 0.01t^4- 0.t^3 }$

Velocity in time T,

$\rm \bold{V_(_t_) = 0.04t^3- 0.6t^2 }$

Velocity after 1 seconds

$\rm \bold{V_(_1_) = 0.02 ft/sec}$

$\rm \bold { 0.04t^3 = 0.6t^2 } }\\\\\rm \bold { t = \frac{3}{2} }$ and when t = 0

the particle will be in rest when $\rm \bold { t = \frac{3}{2} }$ and t = 0.

(D) we need to determine when our velocity function is greater than zero.

Our function is positive when $\rm \bold { -\infty$ and t>3/2.

That is the range in which particle is moving in positive direction.

(E) To find the distance traveled in t =12 seconds, put the value in following equation

$\rm \bold{S_(_t_) = 0.01t^4- 0.t^3 }$

$\rm \bold{S_(_1_2_) = 0.01\times 12^4- 0.\times 12^3 }\\\\\rm \bold{S_(_1_2_) = 172.8 ft }$

(F) As we acceleration is the rate of change in velocity, hence to find the acceleration,

$\rm \bold { a_(_t_) = \frac{dv}{dt} = (0.04t^3- 0.6t^2) } \\\\\rm \bold { a_(_t_) = 0.12t^2- 0.12t }$

The acceleration at t = 1

$\rm \bold { a_(_t_) =0 }$

Therefore we can conclude the particle's acceleration at 1st sec is 0 while after 12 seconds particle move 172.8 ft.

To know more about Law of motion, refer to the link:

link

Answer from: Quest

carbon is found in all living things. carbon atoms move constantly through

living organisms, the oceans, the atmosphere, and the earth’s crust in what is

known as the carbon cycle. the directions taken by carbon atoms through this

cycle are very complicated and can take millions of years to make a full circle.

all animals, from humans to the dinosaurs are part of the carbon cycle. when

animals eat food, they get carbon in the form of carbohydrates and proteins.

in animals, oxygen combines with food in the cells to produce energy for daily

activity and then gives off carbon. the carbon combines with oxygen to form

carbon dioxide (co2) and is released back into the atmosphere as a waste

product when animals breathe and exhale.

from 145-65 million years ago, earth was much

hotter than today and covered with dense,

tropical swamp forests. the trees and other

plants were immense and

provided an endless supply of

food for the giant animals that

roamed the land. somewhere in

the air above one of these

forests, a lone carbon atom has joined

up with two oxygen atoms to form a molecule of co2.

the co2 molecule was sucked into the tiny holes (stomata) on the leaf of a

fern plant and joined with sunlight, chlorophyll and water to make food and

energy in the plant’s cells through photosynthesis. the oxygen (o2) from the

co2 molecule was sent back into the atmosphere; the carbon atom (c) was

detached and used to make a molecule of sugar.

if that carbon atom had been eaten instead, a totally

different situation would have developed. let’s say

that a giant plant-eating dinosaur (herbivore) ate the

fern for breakfast, and swallowed the carbon in the

fern in the form of carbohydrates and proteins. in

the process of respiration, oxygen combined with

the carbohydrates and proteins in the dinosaur’s cells

to provide energy for its daily activity.

Answer from: Quest

true tan speed = ang vel x radius

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Aparticle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s i...

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