A particle 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 − 0.0

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A particle 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 − 0.0

2t3 (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

Physics

1 answer:

serg [3.5K] · Answer 1 · 2026-03-14T02:26:17+03:00

Since you've completed parts a and b, I will tackle part c.
For part C
To respond to this question, we must identify the zeros of the velocity function:
$v(t)=0.04t^3-0.06t^2$
This polynomial can be factored:
$v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)$
Finding the zeros now becomes straightforward since the function equals zero when any factor is zero.
$t^2=0;\\ 0.04t-0.06=0$
By solving these equations, we identify our zeros:
$t_1=0; t_2=\frac{3}{2}$
The particle remains stationary at t=0 and t=3/2.
For part D
We must discover when the velocity function exceeds zero. We will utilize its factored form.
We will assess when each factor is greater than zero and compile the findings in the following table:
$\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}$
From the table, it's evident that our function is positive when $- \infty < t$ and t>3/2.
This indicates the interval during which the particle moves forward.
For part E
The distance traveled can be represented as:
$s(t)=0.01t^4 - 0.02t^3$
We simply substitute t=12 to calculate the total distance traveled:
$s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft$
For part F
Acceleration is defined as the rate at which velocity changes.
We determine acceleration by deriving the velocity function concerning time.
$a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t$
To find the acceleration at 1 second, we substitute t=1s into the previous equation:
$a(1)=0.12-0.12=0$