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:

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Answer from: Quest

magnitude and direction

Answer from: Quest

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