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Paha777 [63]
3 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]3 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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skelet666 [1.2K]

Answer:

DG = 30

Step-by-step explanation:

Given:

DH = 6

DE = 4

EF = 16

Required:

DG

Solution:

DG = DH + HG

DG = 6 + HG

Let's find HG

Given that HE is parallel to the third side of ∆DGF, based on the side-splitter theorem, the other two sides of ∆DGF are divided proportionally.

Therefore,

DH/HG = DE/EF

6/HG = 4/16

Cross multiply

HG*4 = 16*6

HG = 96/4

HG = 24

✔️DG = 6 + HG

DG = 6 + 24

DG = 30

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Brainly for step by step correct answer ❤️
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6×4=24+5=29 inches squared
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Step-by-step explanation:

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I hope this helps!
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