I really need help with my math question if anyone is feeling nice it would be so much to help

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

X^6+3x^3-5 <br> write the expression in quadratic form

Art [367]

Step-by-step explanation:

$v = {x}^{3} \\ {v}^{2} + 3v - 5 = 0 \\$

Solving this equation using the quadratic formula, we get two real solutions :

1.1926 or -4.1926

Now we know the values of v , we can calculate x since x is ∛ v

${x}^{6} + 3 {x}^{3} - 5 = 0$

$x = \sqrt[3]{1.1926} = 1.0605 \\ x = \sqrt[3]{ - 4.1926} = - 1.6125$

3 0

3 years ago

Question 2 (1 point)

Marrrta [24]

Answer:

acute

Step-by-step explanation:

obtuse is over 90°

acute I less that 90°

and right is exactly 90°

6 0

3 years ago

Multiply and Simplify<br> (x-4)(x^2 - 5x - 6)

kirill [66]

Answer:

x^3 -9x^2 +14x +24

Step-by-step explanation:

(x-4)(x^2 - 5x - 6)

Multiply the x by everything in the second term

x * (x^2 - 5x - 6)

x^3 -5x^2 -6x

Multiply the -4 by everything in the second term

-4 * (x^2 - 5x - 6)

-4x^2 +20x +24

Add everything together

I like to line them up vertically

x^3 -5x^2 -6x

-4x^2 +20x +24

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

x^3 -9x^2 +14x +24

4 0

3 years ago

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a family of 4 spend $104 for tickets to a concert all of the tickets were the same price what was the cost of each ticket?

USPshnik [31]

Hello There!

104 / 4 = 26
It was $26 per ticket.

Hope This Helps You!
Good Luck :)

- Hannah ❤

7 0

3 years ago

Read 2 more answers