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Paha777 [63]
4 years ago
8

A car driving on a straight track accelerates at a rate of 3.2 m/s2 for 14 s. If the initial velocity of the car was 5.1 m/s, an

d its initial position was 0 m, what is its final position?
Mathematics
1 answer:
liraira [26]4 years ago
6 0

Answer:

The data we have is:

The acceleration is 3.2 m/s^2 for 14 seconds

Initial velocity = 5.1 m/s

initial position = 0m

Then:

A(t) = 3.2m/s^2

To have the velocity, we integrate over time, and the constant of integration will be equal to the initial velocity.

V(t) = (3.2m/s^2)*t + 5.1 m/s

To have the position equation, we integrate again over time, and now the constant of integration will be the initial position (that is zero)

P(t) = (1/2)*(3.2 m/s^2)*t^2 + 5.1m/s*t

Now, the final position refers to the position when the car stops accelerating, this is at t = 14s.

P(14s) =   (1/2)*(3.2 m/s^2)*(14s)^2 + 5.1m/s*14s = 385m

So the final position is 385 meters ahead the initial position.

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What is the solution to this equation?
Anettt [7]

Answer:

B. a = 5

Step-by-step explanation:

3(a - 4) + 1 = 9 - a

Distribute the 3 into the parenthesis.

3a - 12 + 1 = 9 - a

Add -12 and 1.

3a - 11 = 9 - a

Add a to both sides.

4a - 11 = 9

Add 11 to both sides.

4a = 20

Divide both sides by 4.

a = 5.

Proof:

3(a - 4) + 1 = 9 - a

Substitute variable.

3(5 - 4) + 1 = 9 - 5

Subtract inside parenthesis.

3(1) + 1 = 9 - 5

Multiply 3 and 1.

3 + 1 = 9 - 5

Add 3 and 1.

4 = 9 - 5

Subtract 5 from 9.

4 = 4.

6 0
3 years ago
Given that (-4,1) is on the graph of f(x), find the corresponding point for the function 4f(x)
AVprozaik [17]
F(-4) = 1
4f(-4) = 4

The point of interest is (-4, 4).
8 0
3 years ago
Suppose that the price​ p, in​ dollars, and the number of​ sales, x, of a certain item follow the equation 4 p plus 4 x plus 2 p
andreev551 [17]

Answer:

There is a decrease of 0.75 sales per day

Step-by-step explanation:

Given:-

- The price of item = p

- The number of sales = x

- The relationship between "p" and "x" is given below:

4p+4x+2px=56

Find the rate at which x is changing when x=2, p=6, and dp/dt=1.5 The rate at which x is changing is [ ] sales per day

Take the time derivative (d/dt) of the entire given expression and apply chain rule on d/dt ( 2px ). Since both "p" and "x" are only functions of time "t":

d/dt

4p+4x+2px=56

4*dp/dt + 4*dx/dt + 2*(x*dp/dt + p*dx/dt)=0

Use the given values x=2, p=6, and dp/dt=1.5 to determine dx/dt

4*1.5 + 4*dx/dt + 2*(2*1.5+6*dx/dt)=0

6+4dx/dt + 2*(3+6dx/dt)=0

6+4dx/dt + 6+12dx/dt=0

12+16dx/dt=0

12=-16dx/dt

dx/dt= 12/-16

= -0.75

7 0
4 years ago
Josh has to read a section of his book for his social studies class. For four days, he recorded the number of pages he read and
il63 [147K]

Answer:

3rd is the most

Step-by-step explanation:

divide pages by minutes

1st is .79

2nd is .825

3rd is .88

4th is .72

4 0
3 years ago
URGENT PLEASE HELP: Simplify
Tcecarenko [31]

Answer:

cot (x)

Step-by-step explanation:

cot x

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

cos x  sec x

We know that sec = 1/ cos

cot x

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cos x  1/ cos x

cot x

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

1

cot (x)

3 0
3 years ago
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