math help, any one can help please

Question

A ship made a trip of 231 mi in 14 h. The ship traveled the first 98 mi at a constant rate before increasing its speed by 5 mph. It traveled another 133 mi at the increased speed. Find the rate of the ship for the first 98 mi.

Answer 1

Let's assume the speed of travel for first 98 mi was x mph
Therefore time taken to travel 98 mi @ x mph =(98/x) hr
And, also, the speed of travelling remaining 133 mi must be (x+5) mph
Thus, time taken to travel remaining 133 mi @ (x+5) mph must be = [133/ (x+5)] hr

The total time of the trip of the ship = {(98/x)+[133/ (x+5)] }hr

 

Given that,

{(98/x)+[133/ (x+5)] }=14

=>{(14/x)+[19/ (x+5)] }=2 .............. dividing both sides by 7

=>[14x+70+19x]/{x(x+5)}=2.......... adjusting operators and coefficients

=>33x+70=2x(x+5) ....................adjusting operators and coefficients & multiplying both sides by {x(x+5)}

=>33x+70=2x²+10x ......................adjusting operators and coefficients

=>2x²+10x-33x-70=0 ................ subtracting [33x+70] from both sides

=>2x²-23x-70=0 ................ subtracting [33x+70] from both sides
=>2x²-23x-70=0 

=>x=[23±√{(23)²-(*4*2*[-70])}]/(2*2) =[23±√{529+560}]/4=[23±√{1089}]/4 =[23±33]/4=56/4 or -10/4=14 or -5/2

Since speed cannot be negative in reality, x=14 mph

Therefore  the rate of traveling of the ship for the first 98 mi is 14 mph.

 

Alternative:

Let's assume that the ship travel 98 mi in t hr where t<14 under given conditions

Hence its speed = 98/t mph

Therefore it travelled 133 mi in (14-t) hr at speed 133/(14-t) mph

 Under given condition

98/t+5=133/(14-t)

=>[98+5t]/t= 133/(14-t).......... adjusting operators and coefficients
=>[98+5t](14-t)=133t................ multiplying both sides by t(14-t)

=>1372+70t-98t-5t²=133t.......... adjusting operators and coefficients
=>5t²+133t+98t-70t-1372=0 ................subtracting [1372+70t-98t-5t²] from both sides

=>5t²+161t-1372=0 .......... adjusting operators and coefficients

=>5t²+196t-35t-1372=0 .......... adjusting operators and coefficients

=>t(5t+196)-7(5t+196)=0 .......... adjusting operators and coefficients
=>(5t+196)(t-7)=0 .......... adjusting operators and coefficients
=>t=7 OR t=196/5=39.2>14 and thus does conform with the given conditions

Hence  the ship travel 98 mi in 7 hr  and thus  its speed = 98/7 mph 14 mph

 

Therefore the rate of traveling of the ship for the first 98 mi is 14 mph.

 

Answer 2

This solution relies on the knowledge that v=d/t. Rearranging that equation yields: t=d/v. Let's say that the velocity we are looking for is "v".

 

     98/v +133/(v+5) = 14

 

Multiply through by v(v+5)

 

     98*(v+5) + 133v= 14v(v+5)

 

Distribute:

 

      98v+98*5 +133v = 14v²+70v

 

Combining everything on the RHS to create a quadratic in v

 

      0=14v²-161v-490

 

This yields 2 possible solutions

 

     v=14, -5/2

 

We discard the negative solution as non-physical and find that v = 14 mph.