Question

a train travels at a certain average speed for a distance of 63 km and then travels for a distance of 72 km at an average speed of 6 km/h more than its original speed. if it takes 3 hours to complete total journey, what is the original average speed

Answer 1

Answer:

Step-by-step explanation:

let original average speed be x km/hr.

Let x and (x + 6kmph) represent the speeds traveling 63km and 72km respectively

Question States***total time of trip 3hr. t = D/r

63/x + 72/(x+6) = 3 |Multiplying each term by x(x+6) so as all denominators= 1

63(x+6) + 72x = 3x(x+6)

63x + 6*63 + 72x = 3x^2 + 18x

3x^2 -117x - 378 = 0 3(x^2 - 39x - 126) = 0

factoring

(x+3)(x-42) = 0

(x+3) = 0 |tossing out negative solution

(x-42) = 0

x = 42kmph, speed of train while traveling 63kmph, the original average speed

CHECKING our Answer ***

63km/42kmph + 72km/48kmph = 1.5hr + 1.5hr = 3hr

Answer 2

$\sf\huge{\underline{Answer}}$

$\sf\bold{\boxed{\underline{QUESTION}}}$

A train travels at a certain average speed for a distance of 63 km and then travels for a distance of 72 km at an average speed of 6 km/h more than its original speed. if it takes 3 hours to complete total journey, what is the original average speed?!

$\sf\mathbb{To\: Find}$?!

▶Average Velocity.!!

$<b><h>Solution!!!!$

let the original average speed be 'v(av)', let the distance be s1= 63km and time taken to cover the distance s1 be t1..

i.e,

v(av) = $\frac{s1}{t1}$

t1 = $\frac{s1}{v(av)}$==(1)

while, then he covers distance of s2 = 72km, with a velocity v'(av) =( 6 + v(av) )km\h in time t2.

i.e, t2 = $\frac{s2}{v'(av)}$ = $\frac{s2}{v(av) + 6}$ ==(2)..

according to the question..

t1 + t2 = 3hrs

substituting the values'

$\frac{s1}{v(av)}$ + $\frac{s2}{v(av) + 6}$ = 3

$\frac{63}{v(av)}$ + $\frac{72}{v(av) + 6}$ = 3

$\frac{1}{v(av)}$($\frac{72}{1+6} + 63$) = 3

$\frac{72+63\times 7}{7}$ = 3v(av)

$\frac{441+72}{7}$ = 3v(av)

$\frac{513}{7}$ = 3v(av)

73.28 = 3v(av)

v(av) = $\frac{73.28}{3}$

v(av) = 24.42km\h..

Hence, the original average speed of block is $\sf{\bluebox{24.42km\h\:or\:6.78}}$$ms^{-1}$!!!!!!!