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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.

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Answer:

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

Speed during second part of trip = 53 miles/hour

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 =
x

Distance covered in this part = 250 miles.

Time taken in this trip =
(Distance)/(Speed)=(250\ miles)/(x\ miles/hour)=(250)/(x)\ hours

2) Car driven after rain stopped:

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

Distance covered in this part = 159 miles.

Time taken in this trip =
(Distance)/(Speed)=(159\ miles)/((x+3)\ miles/hour)=(159)/(x+3)\ hours

Total time driven can be given as:


(250)/(x) +(159)/(x+3) = 8

Solving for
x.

Taking LCD.


(250(x+3))/(x(x+3))+(159x)/(x(x+3))=8

Simplifying.


(250x+750+159x)/(x^2+3x)=8


(409x+750)/(x^2+3x)=8

Multiplying both sides by
(x^2+3x)


(x^2+3x)(409x+750)/(x^2+3x)=8(x^2+3x)


409x+750=8x^2+24x

Subtracting both sides by 750.


409x+750-750=8x^2+24x-750


409x=8x^2+24x-750

Subtracting both sides by
409x


409x-409x=8x^2+24x-409x-750


0=8x^2-385x-750

Applying quadratic formula.


x=(-b\pm√(b^2-4ac))/(2a)


x=(-(-385)\pm√((-385)^2-4(8)(-750)))/(2(8))


x=(385\pm415)/(16)


x=(385+415)/(16) and
x=(385-415)/(16)


x=50 and
x=-1.875

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 =
50+3 = 53 miles/hour

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