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 =
Distance covered in this part = 250 miles.
Time taken in this trip =
![(Distance)/(Speed)=(250\ miles)/(x\ miles/hour)=(250)/(x)\ hours](https://img.qammunity.org/2021/formulas/mathematics/middle-school/g8i3phak6may5jzzgaqth5iqu9ao44hqin.png)
2) Car driven after rain stopped:
Speed of the car during this part in miles/hour will be =
![x+3](https://img.qammunity.org/2021/formulas/mathematics/college/b7275fqi5i188dytutfbu5bgt98ijt1i0p.png)
Distance covered in this part = 159 miles.
Time taken in this trip =
![(Distance)/(Speed)=(159\ miles)/((x+3)\ miles/hour)=(159)/(x+3)\ hours](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ix6sz2jgocb8rrbnxn4ottn8224r13udvd.png)
Total time driven can be given as:
![(250)/(x) +(159)/(x+3) = 8](https://img.qammunity.org/2021/formulas/mathematics/middle-school/eake9sh3z48v44m6m91wv4g6zi1pqnna43.png)
Solving for
.
Taking LCD.
![(250(x+3))/(x(x+3))+(159x)/(x(x+3))=8](https://img.qammunity.org/2021/formulas/mathematics/middle-school/oehodlpfygzkjxxvztrifvlz9vds42guub.png)
Simplifying.
![(250x+750+159x)/(x^2+3x)=8](https://img.qammunity.org/2021/formulas/mathematics/middle-school/l168ksf0kue8th3dapmdhvkbeoxqbe3sgd.png)
![(409x+750)/(x^2+3x)=8](https://img.qammunity.org/2021/formulas/mathematics/middle-school/5en6y98wsy9282hy11vbv75sbnznvrij46.png)
Multiplying both sides by
![(x^2+3x)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/rtao59vgd4t9p0m9ttj4dlgboz4sg0stn1.png)
![(x^2+3x)(409x+750)/(x^2+3x)=8(x^2+3x)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/1z9sfvxfm6hniu0owqn37223x36d472lnq.png)
![409x+750=8x^2+24x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/62t022qjj9v26yzz8ou47ttfppyek0dk92.png)
Subtracting both sides by 750.
![409x+750-750=8x^2+24x-750](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vk00kst1xgdfrwyhemofnde2g2vgbchb8l.png)
![409x=8x^2+24x-750](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ontzh623sozkjzdnjdi1q0o1hed1nvs5r3.png)
Subtracting both sides by
![409x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ktupk0639ba9v39bw5f8sny2bq60apfl9l.png)
![409x-409x=8x^2+24x-409x-750](https://img.qammunity.org/2021/formulas/mathematics/middle-school/gt1t9owl9tlpobn4cs4pr5pp5kxv6mhb5l.png)
![0=8x^2-385x-750](https://img.qammunity.org/2021/formulas/mathematics/middle-school/wwre0x4ssa9kv1mvy9a7xdfek1prflq9y3.png)
Applying quadratic formula.
![x=(-b\pm√(b^2-4ac))/(2a)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/g0ca7qogjzpe8ajn601can4liohtrxutl2.png)
![x=(-(-385)\pm√((-385)^2-4(8)(-750)))/(2(8))](https://img.qammunity.org/2021/formulas/mathematics/middle-school/3u6ulgj8wym9z84zqyvgra4a4hqp0r3f80.png)
![x=(385\pm415)/(16)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/bmmwaglakdatyl0e9hvovalj1x8eal8gfa.png)
and
![x=(385-415)/(16)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/wbopydqmu23agnmnz6oq3ffrsbbg7cuhjd.png)
∴
and
![x=-1.875](https://img.qammunity.org/2021/formulas/mathematics/middle-school/h89ayid0yqmribgd701z9fepufex40z5mg.png)
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 =
= 53 miles/hour