a boat covers a certain distance downstream in 2 hours. it covers the same distance upstream in 2 and a half hours. the speed of the boat in still water is 18 km hr. find the speed of the water also find the distance covered by the boat
Answers
Answer:
Wboat covers a certain distance downstream in 2 hours.
it covers the same distance upstream in 2.5 hours.
The speed of the boat i 18 km/hr.
Find the speed of the water.
let c = the speed of the water
then
(18-c) effective speed upstream
and
(18+c) = effective speed down
Write a distance equation. dist = time * speed
down dist = up dist
2(18+c) = 2.5(18-c)
36 + 2c = 45 -2.5c
2c + 2.5c = 45 - 36
4.5c = 9
c = 9/4.5
c = 2 km/hr is the speed of the current
:
Also find the distance covered by the boat
2(18+2) = 40 km is the distance
confirm this upstream
2.5(18-2) = 40 km
Answer:
Let the speed of water be x.
\sf{2 \ and \ half \ hour=2.5 \ hr}2 and half hour=2.5 hr
\boxed{\sf{Distance=Time\times \ Speed}}
Distance=Time× Speed
\sf{According \ to \ first \ condition. }According to first condition.
\sf{Distance_{downstream}=2\times(18+x)...(1)}Distance
downstream
=2×(18+x)...(1)
\sf{According \ to \ the \ second \ condition. }According to the second condition.
\sf{Distance_{upstream}=2.5\times(18-x)...(2)}Distance
upstream
=2.5×(18−x)...(2)
\sf{But,}But,
\sf{Distance_{downstream}=Distance_{upstream}...Given}Distance
downstream
=Distance
upstream
...Given
\sf{...from \ (1) \ and \ (2)}...from (1) and (2)
\sf{2\times(18+x)=2.5\times(18-x)}2×(18+x)=2.5×(18−x)
\sf{Multiply \ both \ sides \ by \ 2, \ we \ get}Multiply both sides by 2, we get
\sf{4\times(18+x)=5\times(18-x)}4×(18+x)=5×(18−x)
\sf{\therefore{72+4x=90-5x}}∴72+4x=90−5x
\sf{\therefore{4x+5x=90-72}}∴4x+5x=90−72
\sf{\therefore{9x=18}}∴9x=18
\sf{\therefore{x=\frac{18}{2}}}∴x=
2
18
\boxed{\sf{\therefore{x=2}}}
∴x=2
\sf{\therefore{Speed \ of \ water=2 \ km \ hr^{-1}}}∴Speed of water=2 km hr
−1
\sf{Distance_{downstream}=2\times(18+2)}Distance
downstream
=2×(18+2)
\sf{\therefore{Distance_{downstream}=2\times20}}∴Distance
downstream
=2×20
\sf{\therefore{Distance_{downstream}=40 \ km \ hr^{-1}}}∴Distance
downstream
=40 km hr
−1
\sf\purple{\tt{\therefore{The \ speed \ of \ water \ is \ 2 \ km \ hr^{-1} \ and}}}∴The speed of water is 2 km hr
−1
and
\sf\purple{\tt{distance \ is \ 40 \ km}}distance is 40 km
Step-by-step explanation: