Please help 30 points will mark brainliest.

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago

The intensity of light bulb measured in watts per square meter (w/m^2), varies inversely as the square of the distance d from th

krek1111 [17]

Answer:

I(7) = 22.96 W/m²

Step-by-step explanation:

Let intensity of light, be I(d) where d is distance. There must be a proportionality factor K such as

I(d) = K /d²

Now for d = 5 m I(d) = K* 1/d²

then 45 * 25 = K

K = 1125

To find the intensity when the distance from the light bulb is 7 m away

I(7) = K / 49

I(7) = 1125/49 W/m²

I(7) = 22.96 W/m²

6 0

3 years ago

I need help with eight divided by three

mezya [45]

2.6666666667. It is endless, kinda.

4 0

3 years ago

Read 2 more answers

Which situation represents a proportional relationship?

lyudmila [28]

Answer:

D. The cost of purchasing avocados for $1.75 per pound plus a shipping fee of $0.16 per pound.

Step-by-step explanation:

A linear relationship is of the form $y=mx+b$ , where, $m$ is the unit rate and $b$ is the fixed value (constant).

For a proportional relationship, the value of $b=0$ and thus it is of the form $y=mx$

Let us check each option and express it in the form above.

Option A:

Given:

Unit rate of purchasing a basket of apples, $m=\$ 1.25$

Fixed price for the basket, $b=\$ 5$

Since, $b \ne 0$ , therefore, it is not a proportional relationship.

Option B:

Given:

Unit rate of purchasing a banana, $m=\$ 5$

Fixed price for the box, $b=\$ 3$

Since, $b \ne 0$ , therefore, it is not a proportional relationship.

Option C:

Given:

Unit rate of purchasing a lime, $m=\$ 0.65$

Discount from total cost, $b=\$ 1$

Since, $b \ne 0$ , therefore, it is not a proportional relationship.

Option D:

Given:

Unit rate of purchasing an avocado = $\$ 1.75$

Unit rate of shipping = $\$ 0.16$

Therefore, total cost per pound is the sum of the unit rates of purchasing and shipping. So,

Total cost of avocados per pound, $m=1.75+0.16=\$ 1.91$

There is no fixed cost on this. So, $b=0$

Since, $b = 0$ , therefore, it is a proportional relationship.

Therefore, the correct option is D.

6 0

3 years ago

Find range of function y=1/2x+3 when domain =-2,0,2,4

eduard

The Answer Is Negative 6

8 0

3 years ago