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allsm [11]
3 years ago
5

Two trains are headed towards each other on the same track unbeknownst to the engineers. One departs San Francisco. Its average

speed is 55 miles per hour. The 2nd departs Seattle at the same time that the first one departs San Francisco. Its average speed is 65 miles per hour. The length of track separating the two locations is 860 miles. How many minutes (from departure time) before the two trains collide?
Physics
1 answer:
aalyn [17]3 years ago
7 0

Answer:

7,166 hrs =430  minutes

Explanation:

Since both train are on the same track, going one towards the other, the relative speed is the addition of both, then the time they need to meet, and consistently crash, is the time that (65mph + 55 mph)=120mph need to travel the total distance of 860 miles, of course in this case one part is traveled by the first train and the rest by the other. Then to find the time we use a three rule

1 h --->120mi

X ---->860mi, then X=(860 mi* 1h)/120 mi = 43/6 hrs= 7,16666 hrs, turning this into minutes need that we notice 1h=60min, then 43/6 hrs *60 min/hrs = 430 minutes.

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and where I put a negative sign in front of the acceleration of the second player, since he's moving in the opposite direction of the first player.

The time t at which the two players collide is the time t at which x_1 = x_2, therefore:
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t= \sqrt{ \frac{2d}{a_1+a_2} }= \sqrt{ \frac{2 \cdot 36 m}{0.58 m/s^2+0.42 m/s^2} }=8.5 s

b) We can use the equation of x_1(t) to find how far the first player run in t=8.5 s:
x_1(t)= \frac{1}{2}a_1 t^2= \frac{1}{2}(0.58 m/s^2)(8.5 s)^2=21.0 m
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