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OLEGan [10]
3 years ago
12

2 cars raced at a track. the faster car traveled 20mph faster than the slower car. in the time that the slower car traveled 165

miles, the faster car traveled 225 miles. if the speeds of the cars remained constant, how fast did the slower car travel during the race.
Mathematics
1 answer:
icang [17]3 years ago
3 0

Answer:

The speed of slower car is 55 miles per hour.

Step-by-step explanation:

Given as :

The speed of slower car = s_2 = s  mph

The speed of faster car = s_1 = ( s + 20 ) mph

The distance cover by slower car = d_2 = 165 miles

The distance cover by faster car = d_1 = 225 miles

The time taken by both cars for travelling = t hours

The speed of the cars remains constant

Now, <u>According to question</u>

∵ Time = \dfrac{\textrm Distance}{\textrm Speed}

So,<u> For slower car</u>

t = \dfrac{d_2}{s_2}

Or, t = \dfrac{165}{s}           ............1

So, For faster car

t = \dfrac{d_1}{s_1}

Or, t = \dfrac{225}{s+20}           ............2

Now, equating both the equations

I.e  \dfrac{225}{s+20} = \dfrac{165}{s}  

<u>By cross multiplying</u>

Or, 225 × s = 165 × (s + 20)

Or, 225 s = 165 s + 3300

Or, 225 s - 165 s = 3300

Or, 60 s = 3300

∴  s = \dfrac{3300}{60}

I.e s = 55 miles per hour

So , The speed of slower car = s_2 = s = 55 miles per hour

Hence , The speed of slower car is 55 miles per hour.  Answer

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In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En
mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

\frac{dy}{dx} = 3t^{2} - 2t - 27

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

3t^{2} - 2t - 27 = 0

So

a = 3, b = -2, c = -27

Then

\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328

So

t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35

t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55

y = t^{2} - 4 = (3.35)^2 - 4 = 7.22

The first point is (27.55, 7.22)

t = -2.685

x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3

y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21

The second point is (-11.3, 3.21).

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