Please answer get this right please i’ll do anything

You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in

emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

$j(t)=95t-9t^2$

$j'(t)=95-18t$

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

-18t=-95

$t=\frac{95}{18}$

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

$j(5.28)=95(5.28)-9(5.28)^{2}$

$j(5.28)=250.69ft$

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

$j(t)=95t-9t^{2}$

$j'(t)=95-18t$

j"(t)=-18

and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

5 0

3 years ago

The

stepan [7]

Answer:

$Area = 129.5m^2$

$Perimeter = 48m$

Step-by-step explanation:

Given

See attachment

Required

Determine the area and the perimeter of the garden

Calculating Area

First, we calculate the $area\ of\ the\ rectangle$

$A_1 = L * B$

Where:

$L = 20-(3.5 +3.5)$

$B = 7$

So:

$A_1 = (20 - (3.5 + 3.5)) * 7$

$A_1 = (20 - 7) * 7$

$A_1 = 13 * 7$

$A_1 = 91$

Next, we calculate the area of the two semi-circles.

Two semi-circles = One Circle

So:

$A_2 = \pi r^2$

Where

$r = \frac{7}{2}$

$A_2 = \frac{22}{7} * (\frac{7}{2})^2$

$A_2 = \frac{22}{7} * \frac{49}{4}$

$A_2 = \frac{22}{1} * \frac{7}{4}$

$A_2 = \frac{22*7}{4}$

$A_2 = \frac{154}{4}$

$A_2 = 38.5$

Area of the garden is

$Area = A_1 + A_2$

$Area = 91 + 38.5$

$Area = 129.5m^2$

Calculating Perimeter

First, we calculate the perimeter of the rectangle

But in this case, we only consider the length because the widths have been covered by the semicircles

$P_1 = 2 * L$

Where:

$L = 20-(3.5 +3.5)$

So:

$P_1 =2 * (20-(3.5 +3.5))$

$P_1 =2 * (20-7)$

$P_1 =2 * 13$

$P_1 =26$

Next, we calculate the perimeter of the two semi-circles.

Two semi-circles = One Circle

So:

$P_2 = 2\pi r$

Where

$r = \frac{7}{2}$

$P_2 = 2 * \frac{22}{7} * \frac{7}{2}$

$P_2 = \frac{2 * 22 * 7}{7 * 2}$

$P_2 = \frac{308}{14}$

$P_2 = 22$

Perimeter of the garden is

$Perimeter = P_1 + P_2$

$Perimeter = 26 + 22$

$Perimeter = 48m$

8 0

3 years ago

2. The second term of an arithmetic sequence is -5, and the third term is 12. What is the first term?

kondaur [170]

Answer: -12

Step-by-step explanation:

To get from -5 to 12, you'd add 7

So to get to the first, subtract 7 from -5

4 0

3 years ago

Raju, ramu and razi can do a piece of work in 20, 30 and 60 days respectively depending on their capacity of doing work. If raju

jok3333 [9.3K]

If Raju is assisted by Ramu and Razi every third day, then the number of days in which Raju will complete the work is 15 days.

<h3>What is Division?</h3>

The division is one of the four fundamental arithmetic operations, which tells us how the numbers are combined to form a new one.

Let the total unit of work be 60 units.

Given that Raju, Ramu, and Razi can do a piece of work in 20, 30, and 60 days respectively. Therefore, the rate at which three of them works can be written as,

Rate = Total Work/ Time

Rate of Raju = 60 units /20 days = 3 units per day

Rate of Ramu = 60 units /30 days = 2 units per day

Rate of Razi = 60 units /60 days = 1 units per day

The work completed on the first day will be,

Work completed = Rate × Time

= Rate of Raju × 1 day

= 3 units per day × 1 day

= 3 units

The work completed on the second day will be,

Work completed = Rate × Time

= Rate of Raju × 1 day

= 3 units per day × 1 day

= 3 units

The work completed on the third day will be,

Work completed = Rate × Time

= (Rate of Raju+ Rate of Ramu + Rate of Razi) × 1 day

= (3+2+1) units per day × 1 day

= 6 units

Therefore, every three days, 12 units every three days of work will be completed. Therefore, the number of times three of them need to work together is,

Number of times = 60 units / 12 units every three days = 5

Further, the number of days if they meet 5 times is,

Number of days = 3 × 5 = 15

Hence, If Raju is assisted by Ramu and Razi every third day, then the number of days in which Raju will complete the work is 15 days.

Learn more about Division here:

brainly.com/question/369266

#SPJ4

8 0

2 years ago

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