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muminat
3 years ago
12

two cars travel in opposite directions, starting from the same place at the same time. One travels at an average rate of 48 mile

s per hour and the other averaged at 55 miles per hour. in how many hours will the be 618 miles apart?
Mathematics
1 answer:
asambeis [7]3 years ago
6 0
So in 1 hour they will be 48+55=103 miles apart
how many hours will it be in 618
103 times x hours=618
divide both sides by 103
x=618/103
x=6

the answer is 6 hours
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In quadratic drag problem, the deceleration is proportional to the square of velocity
Mars2501 [29]
Part A

Given that a= \frac{dv}{dt} =-kv^2

Then, 

\int dv= -kv^2\int dt \\  \\ \Rightarrow v(t)=-kv^2t+c

For v(0)=v_0, then

v(0)=-kv^2(0)+c=v_0 \\  \\ \Rightarrow c=v_0

Thus, v(t)=-kv(t)^2t+v_0

For v(t)= \frac{1}{2} v_0, we have

\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\  \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\  \\ \Rightarrow kv_0t=2 \\  \\ \Rightarrow t= \frac{2}{kv_0}


Part B

Recall that from part A, 

v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\  \\ \Rightarrow dx=-kv^2tdt+v_0dt \\  \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\  \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a

Now, at initial position, t = 0 and v=v_0, thus we have

x=a

and when the velocity drops to half its value, v= \frac{1}{2} v_0 and t= \frac{2}{kv_0}

Thus,

x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\  \\ =- \frac{1}{2k} + \frac{2}{k} +a

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

- \frac{1}{2k} + \frac{2}{k} +a-a \\  \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}
7 0
3 years ago
If the line formed by equation 1 is parallel to the line formed by equation 2, fill in the missing value below.
madreJ [45]

Answer:

y =-\frac{3}{2}x + 6

Step-by-step explanation:

Given

y = []x + 6 --- equation 1

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Required

Fill in the blank

A linear equation is represented as:

y = mx + b

Where m represents the slope

If (1) and (2) are perpendicular, then:

m_1 = -\frac{1}{m_2}

In y = \frac{2}{3}x + 10 --- equation 2

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So, we have:

m_1 = -\frac{1}{m_2}

m_1 = -\frac{1}{2/3}

m_1 = -\frac{3}{2}

Hence, the complete equation is:

y =-\frac{3}{2}x + 6

7 0
3 years ago
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