How to use a number line to count to 580 to 994

The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

3 years ago

How to solve these equations?

Scorpion4ik [409]

1) / sin x / = y ≥ 0 ;
We solve the equation $y^{2} +y - 2 = 0 ;
$
a = 1 ; b = 1 ; c = - 2 ;
$b^{2} - 4ac = 9 ;$
y1 = ( - 1 + 3 ) / 2 = 1 ≥ 0 ( correct ) ;
y2 = ( - 1 - 3 ) / 2 = - 2 ≥ 0 ( false ) ;
Then, / sin x / = 1 ;
sinx = + 1 or sin x = - 1 ;
x ∈ { 90 } U { 270 } ;
x ∈ {90 , 270}.

8 0

3 years ago

Uses of scale factor in our daily life

kumpel [21]

Answer:

Scale factor can be used in almost all situations. Scale factor is used in math and can be used when constructing any mini model of a real life design. Scale factor is often utilized in cooking recipes/procedures.

Step-by-step explanation:

5 0

4 years ago

Finally, suppose you decide not to finance (get a loan for) the car and to

Alexxandr [17]

<u>Supposing a cost of $30,000</u>, and using proportions, it is found that you would have to save $1,083.30 each month.

--------------------------

We suppose a cost of $30,000.

You already have $4,000 saved.
30 - 4 = 26, thus, you would need to save $26,000 in 2 years = 24 months.
How much per month?

We can solve it using proportions, by the following rule of three:

1 month - $x

24 months - $26,000

Applying cross multiplication:

$24x = 26000$

$x = \frac{26000}{24}$

$x = 1083.3$

You would have to save $1,083.30 each month.

A similar problem is given at brainly.com/question/24112433

7 0

3 years ago

Will give brainliest foe correct answer

Dima020 [189]

Answer:

true

Step-by-step explanation:

5 0

3 years ago