trains AandB leave station at same time. train A travels in west direction and train Btravels in north direction. trainA is 2km+h slower than train B. after 5 hrs,train are 290 km apsrt.find the speed of both trains..
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Let s = the speed of the northbound train
Then
(s+5) = the speed of the westbound train
:
This is a right triangle problem: a^2 + b^2 = c^2
The distance between the trains is the hypotenuse
dist = speed * time
The time is 2 hrs, so we have
a = 2s; northbound train distance
b = 2(s+5) = (2s+10); westbound distance
c = 50; distance between the two trains
:
(2s)^2 + (2s+10)^2 = 50^2
4s^2 + 4s^2 + 40s + 100 = 2500
Arrange as a quadratic equation4s^2 + 4s^2 + 40s + 100 - 2500 = 0
8s^2 + 40s - 2400 = 0
:
Simplify, divide by 8:
s^2 + 5s - 300 = 0
:
Factors to
(s - 15)(s + 20) = 0
:
The positive solution is what we want here
s = 15 mph is the speed of the northbound train
then
5 + 15 = 20 mph is the speed of the westbound train
:
:
Check this; find the distance (d) between the trains using these distances
Northbound traveled 2(15) = 30 mi
Westbound traveled 2(20) = 40 mi
d = 50
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Then
(s+5) = the speed of the westbound train
:
This is a right triangle problem: a^2 + b^2 = c^2
The distance between the trains is the hypotenuse
dist = speed * time
The time is 2 hrs, so we have
a = 2s; northbound train distance
b = 2(s+5) = (2s+10); westbound distance
c = 50; distance between the two trains
:
(2s)^2 + (2s+10)^2 = 50^2
4s^2 + 4s^2 + 40s + 100 = 2500
Arrange as a quadratic equation4s^2 + 4s^2 + 40s + 100 - 2500 = 0
8s^2 + 40s - 2400 = 0
:
Simplify, divide by 8:
s^2 + 5s - 300 = 0
:
Factors to
(s - 15)(s + 20) = 0
:
The positive solution is what we want here
s = 15 mph is the speed of the northbound train
then
5 + 15 = 20 mph is the speed of the westbound train
:
:
Check this; find the distance (d) between the trains using these distances
Northbound traveled 2(15) = 30 mi
Westbound traveled 2(20) = 40 mi
d = 50
plz mark me as brain list
Answered by
0
Answer:
D=50 pls mark me as a brainlist
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