Question

A train X starts from "A" at 5pm and reaches "D" at 10 pm.Another train Y starts from from "D" at 7pm and reaches "A" at 11pm .At what time train will meet?

Answer 1

doesnt that depend on other criteria?

Answer 2

the assumptions to be made are that the travel from A to D is in a straight; and that by "meet" it means that the trains dont actually crash into each other but rather are on parallel tracks and pass each other

Answer 3

determine the speed of each train for starters

Answer 4

good morning sensei!

Answer 5

howdy :)

Answer 6

we can do this yes? you should have seen that other circle problem yesterday. wow!

Answer 7

Yes amistre64 your assumptions are correct

Answer 8

this is ones fairly simple ;)

Answer 9

since the distance is arbitrary; lets make another assumption; A and D are 5 miles apart, or simply insert a distance variable of your choosing into the equations

Answer 10

train X travels at a speed of 5 miles per 5 hours

and train Y travels at a speed of 5 miles per 4 hours

Answer 11

train X has a 2 hour head start right?

Answer 12

since train X has a speed of 5/5 mph and a head start of 2 hours (2 miles) we can formulate an equation: tX = t+2

train Y starts the journey at a constant speed of 5/4 mph at that time soo: tY = (5/4) x

when does tY = tX?

Answer 13

(5/4)t would be better notation to conform with the other one :)

but does this make sense to you?

Answer 14

tY = tX when:

t+2 = (5/4)t ; subtract one side from the other

0 = (1/4)t -2 ; solve for t

t = 8

im gonna have to dbl chk my thinking to make sure

Answer 15

Common time interval from 7 to 10 = 3hrs
Tx : Ty = 5 :4
Divide 3hrs of interval in ratio 5 :4
For Y , time of meeting will be (3*4/9) = 1 hr 20 mints
Y started at 7 pm that is they will meet at 8:20

Answer 16

see; that was easy enough :)

Answer 17

We can take it other way round also.

Answer 18

Tell your teacher the question is faulty until he/she defines \(g_1\) and \(g_2\) with the following distance functions:
\[f_1(t)=\left\{\begin{array}{c|c}
0 & t < 5 \\
g_1(t) & t \geq 5 \\
d & t > 10
\end{array}\right.
\]\[
f_2(t)=\left\{\begin{array}{c|c}
d & t <7 \\
d - g_2(t) & 7 \leq t \leq 11 \\
0 & t > 11
\end{array}\right.
\]

Answer 19

r t = d
Let x and y be the speed, unit length per hour, that is required to move 5 unit lengths of distance respectively.
{x t = 5, y t = 5}

X takes 5 hours and Y takes 4 hours to cover the 5 unit distance.
{x 5 = 5, y 4 = 5} -> {x = 1, y = 5/4}

At 7 pm, X has 3 unit lengths to go and Y is just starting to move. Solve for the time it takes to move 3 unit lengths using their combined speeds.
(1 + 5/4) t = 3, t = 4/3, 7 pm + 1 : 20 = 8 : 20 pm