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grigory [225]
3 years ago
7

A car is traveling in a race. The car went from the initial velocity of 35 m/s to the final velocity of 65m/s in 5 seconds what

is the acceleration?
Mathematics
2 answers:
Lerok [7]3 years ago
5 0

Answer:

6m/s2

Step-by-step explanation:

artcher [175]3 years ago
3 0

The acceleration is defined as the ratio between the change in velocity and the time elapsed to perform such a change.

These "changes" are indicated with the capital greek letter delta, \Delta, and when you write \Delta x you mean the difference between the finial and the inital values of the variable x:

\Delta x = x_{\text{fin}} - x_{\text{init}}

So, the acceleration is defined as

a = \dfrac{\Delta v}{\Delta t} = \dfrac{v_{\text{fin}} - v_{\text{init}}}{t_{\text{fin}} - t_{\text{init}}}

In this case, the initial velocity is 35, the final velocity is 65. Assuming we start the clock at the beginning of the observation, the inital time is 0 and the final time is 5. So, we have

a = \dfrac{65-35}{5-0} = \dfrac{30}{5} = 6m/s^2

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Answer:

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Step-by-step explanation:

<u>To solve quadratic systems,we always substitute one variable in terms if the other and then solve the equation.</u>

x + 2y = 6                                 ---------------(1)

y - 5 = (x-2)^{2}         ---------------(2)

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Substitute (3) in (1) ,

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(a + b)^{2} =a^{2} + 2ab + b^{2}

x + 2( x^{2} - 4x + 4 + 5 ) = 6

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x = \frac{(-b) + \sqrt{(-b)^{2}-4 \times ac }  }{2 \times a}  -----------(5)

According to equation (5),solution of (4) is

x =  \frac{7 + \sqrt{(-7)^{2}-4 \times 24 }  }{2 \times 2}

x =  \frac{7+\sqrt{49-96}}{4}

x = \frac{7+\sqrt{47}\times i }{4}

 

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Step-by-step explanation:

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