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Dimas [21]
3 years ago
11

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

' (t) is its acceleration.
x(t) = t3 − 6t2 + 9t − 5, 0 ≤ t ≤ 10

(a) Find the velocity and acceleration of the particle.

x' (t) =
x'' (t) =


(b) Find the open t-intervals on which the particle is moving to the right. (Enter your answer using interval notation.)



(c) Find the velocity of the particle when the acceleration is 0.
Mathematics
1 answer:
vodka [1.7K]3 years ago
5 0

Answer:

a) v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

b)  0

c) a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9

a(t) = x''(t) = v'(t) =6t-12

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

v(t) = 3t^2 -12t +9

We can factorize this function as v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is 0\leq t\leq 10 we need to analyze the following intervals:

0< t

For this case if we select two values let's say 0.25 and 0.5 we see that

v(0.25) =6.1875, v(0.5)=3.75

And we see that for a=0.5 >0.25=b we have that f(b)>f(a) so then the function is decreasing on this case.  

1

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

2

We see that the function would be increasing.

3

For this interval we will see that for any two points a,b with a>b we have f(a)>f(b) for example let's say a=3 and b =4

f(a=3) =0 , f(b=4) =9 , f(b)>f(a)

The particle is moving to the right then the velocity is positive so then the answer for this case is: 0

Part c

a(t) = x''(t) = v'(t) =6t-12

When the acceleration is 0 we have:

6t-12=0, t =2

And if we replace t=2 in the velocity function we got:

v(t) = 3(2)^2 -12(2) +9=-3

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Sveta_85 [38]

Not an expertise on infinite sums but the most straightforward explanation is that infinity isn't a number.

Let's see if there are anything we missed:

 ∞

 Σ  2^n=1+2+4+8+16+...

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We multiply (2-1) on both sides:

       ∞

(2-1)  Σ  2^n=(2-1)1+2+4+8+16+...

      n=0

And we expand;

 ∞

 Σ  2^n=(2+4+8+16+32+...)-(1+2+4+8+16+...)

n=0

But now, imagine that the expression 1+2+4+8+16+... have the last term of 2^n, where n is infinity, then the expression of 2+4+8+16+32+... must have the last term of 2(2^n), then if we cancel out the term, we are still missing one more term to write:

 ∞

 Σ  2^n=-1+2(2^n)

n=0

If n is infinity, then 2^n must also be infinity. So technically, this goes back to infinity.

Although we set a finite term for both expressions, the further we list the terms, they will sooner or later approach infinity.

Yep, this shows how weird the infinity sign is.

5 0
2 years ago
Find the difference between and 8/15 - -(2/3). Show all calculations in your final answer please!! :)
sasho [114]

Answer:

\boxed{\dfrac{34}{15}}

Step-by-step explanation:

\text{Use }\LaTeX (ノ◕ヮ◕)ノ*:・゚✧

\dfrac{8}{5} - - \left(\dfrac{2}{3}\right)

\dfrac{8}{5} - \left[-\dfrac{2}{3}\right]

\dfrac{8}{5} +\dfrac{2}{3}

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\dfrac{24}{15} +\dfrac{10}{15}

\boxed{\dfrac{34}{15}}

3 0
3 years ago
Read 2 more answers
A softball base is 15 x 15 inches long and wide. What is the length of the diagonal?
Finger [1]

Answer:

15 sqrt 2

Step-by-step explanation:

Use Pythagorean Theorem

a^2 + b^2 = c^2

15^2 + 15^2 = c^2

225 + 225 = 450

Find prime factors of the sq rt of 450

450 = 2 * 3 * 3 * 5 * 5

15 sqrt 2

4 0
2 years ago
Quizizz 10 pts each help please WITH PIC brainiest guaranteed
lawyer [7]

Answer:

2.-2m^4-6m^2+4m+9

3.-2m^4-6m^2-4m+9

Step-by-step explanation:

<em>In general we can write a polynomial in standard form as </em>

ax^n+bx^n^-^1+cx^n^-^2+...+px+q

<em>Given</em> 4m-2m^4-6m^2-4m+9

<em>Combine the like terms: 4m and -4m</em>

<em>4m-4m=0</em>

<em>We have 4m-4m=0</em>

<em>So, write the remaining terms</em>

-2m^4-6m^2+9+0

= -2m^4-6m^2+9

<em>This is in decreasing order of powers.</em>

<em>Hence the answer is the standard form is</em>

-2m^3-6m^2+9

<em>But in the given options, you can choose option 2 and option 3 are in standard form.</em>

<em>Because they are in decreasing order of powers.</em>

<em>In other two options, the constants term is first and the highest power term is at the last. So, they are not in standard form.</em>

<em>-2m^4-6m^2+4m+9</em>

<em>-2m^4-6m^2-4m+9</em>

<em>I hope this helps you.</em>

<em>And please comment if I need to do corrections.</em>

<em>Please let me know if you have any questions.</em>

8 0
3 years ago
Brainliest + Points!<br><br> Please explain !
Morgarella [4.7K]

Answer:

A) -2/5

Step-by-step explanation:

Note that for a slope, the slope of the line is found by using the equation:

slope = (rise/run).

In this case, find two points. For example, i will use:

(0, 2) & (5, 0). (x = 0, y = 2) ; (x = 5, y = 0)

Note that "rise" = y, and "run" = x. Plug in the corresponding numbers for the corresponding points.

(2 - 0)/(0 - 5) = slope

Simplify

(2)/(-5) = slope

-(2/5) is your answer.

Note that it does not matter where the negative sign is placed.

~

6 0
3 years ago
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