Greg drove at a constant speed in a rainstorm for 250 miles. He took a​ break, and the rain stopped He then drove 159 miles at a speed that was 3 miles per hour faster than his previous speed. If he drove for 8 ​hours, find the​ car's speed for each part of the trip.

Respuesta :

Answer:

Speed during first part of the trip = 50 miles/hour

Speed during second part of trip  = 53 miles/hour

Step-by-step explanation:

Given:

Distance driven in rainstorm = 250 miles

Distance driven after rain stopped = 159 miles

Speed driven after rain stopped is 3 miles faster than the speed driven in rainstorm.

Total time driven = 8 hours

To find the speed of the car at each part of the trip.

Solution:

There are two parts of the trip.

1) Car driven in rainstorm:

Let the speed of the car during this part in miles/hour be = [tex]x[/tex]

Distance covered in this part = 250 miles.

Time taken in this trip = [tex]\frac{Distance}{Speed}=\frac{250\ miles}{x\ miles/hour}=\frac{250}{x}\ hours[/tex]

2) Car driven after rain stopped:

Speed of the car during this part in miles/hour will be = [tex]x+3[/tex]

Distance covered in this part = 159 miles.

Time taken in this trip = [tex]\frac{Distance}{Speed}=\frac{159\ miles}{(x+3)\ miles/hour}=\frac{159}{x+3}\ hours[/tex]

Total time driven can be given as:

[tex]\frac{250}{x} +\frac{159}{x+3} = 8[/tex]

Solving for [tex]x[/tex].

Taking LCD.

[tex]\frac{250(x+3)}{x(x+3)}+\frac{159x}{x(x+3)}=8[/tex]

Simplifying.

[tex]\frac{250x+750+159x}{x^2+3x}=8[/tex]

[tex]\frac{409x+750}{x^2+3x}=8[/tex]

Multiplying both sides by [tex](x^2+3x)[/tex]

[tex](x^2+3x)\frac{409x+750}{x^2+3x}=8(x^2+3x)[/tex]

[tex]409x+750=8x^2+24x[/tex]

Subtracting both sides by 750.

[tex]409x+750-750=8x^2+24x-750[/tex]

[tex]409x=8x^2+24x-750[/tex]

Subtracting both sides by [tex]409x[/tex]

[tex]409x-409x=8x^2+24x-409x-750[/tex]

[tex]0=8x^2-385x-750[/tex]

Applying quadratic formula.

[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

[tex]x=\frac{-(-385)\pm\sqrt{(-385)^2-4(8)(-750)}}{2(8)}[/tex]

[tex]x=\frac{385\pm415}{16}[/tex]

[tex]x=\frac{385+415}{16}[/tex]     and  [tex]x=\frac{385-415}{16}[/tex]

∴ [tex]x=50[/tex]     and [tex]x=-1.875[/tex]

Since speed cannot be taken as negative, so our solution will be 50 miles per hour.

Speed during first part of the trip = 50 miles/hour

Speed during second part of trip = [tex]50+3[/tex] = 53 miles/hour