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raketka [301]
3 years ago
9

A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave

ling on the same track, but traveling due south with a constant speed of 136 m/s. At what position will the trains collide? Round to the nearest whole number.
Physics
1 answer:
damaskus [11]3 years ago
7 0

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance x m by the time of collision, then train B would have traveled (2881 - x) m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

t_{A} = t_{B}

t_{A} is the time spent by train A

t_{B} is the time spent by train B

From,

Velocity = \frac{Distance }{Time }\\

Time = \frac{Distance}{Velocity}

Since the time spent by the two trains is equal,

Then,

\frac{Distance_{A} }{Velocity_{A} }  = \frac{Distance_{B} }{Velocity_{B} }

{Distance_{A} = x m

{Distance_{B} = 2881 - x m

{Velocity_{A} = 100 m/s

{Velocity_{B} = 136 m/s

Hence,

\frac{x}{100} = \frac{2881 - x}{136}

136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\

x≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

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