Calculus QuestionA train is travelling west at a speed of 18 m/s on a set of tracks that is 15 m below an overpass. I am driving on this overpass going north 25 m/s. At 8:00, the train was below...

Calculus Question

A train is travelling west at a speed of 18 m/s on a set of tracks that is 15 m below an overpass. I am driving on this overpass going north 25 m/s. At 8:00, the train was below the overpass, but I was 20 m from a spot on the overpass that was directly over the train. When will my car and the train be closest together?

Asked on by utilityfan

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txmedteach | High School Teacher | (Level 3) Associate Educator

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If you diagram this out, your distance to the train is based on three separate distances: the height of the overpass (h = 15 m), your North-South distance from the point, and the train's West-East distance from the point.

The easiest way to express the train's and your locations is to set up a coordinate system. Let's let the origin of our coordinate system be the spot on the overpass directly above the train at the initial time (t=0). Let's have the West-East direction be the x-axis and let the North South direction be the y-axis. We won't concern ourselves with the 15 m heigh of the overpass for now because we can calculate the minimum distance based on the 2-D case first.

Your 2-D graph at t = 0 based on the above defintion of our coordinate axes will be the following:

The red dot will be the initial position of the train, and the blue dot will be your initial position.

Based on this coordinate system and our velocities, we can come up with two functions that give the location of the train and your car for any time. Let's let `T(t)` = train's location on the x-axis and C(t) = Your car's location on the y-axis.

`T(t) =-18t`

`C(t) = 25t-20`

These equations are based on the initial position (the constant term) and the velocity (the term containing "t").

Clearly, at any time except for t = 0, you can construct a right triangle with the location of your car, the train, and the origin of our axes. This fact will give us the distance between the car and the train (recall, this is only the 2-D distance, not the 3-D distance the problem asks for).

So, let's get a formula for the projected distance. Recall the pythagorean theorem with `D_(2D)(t)` being the 2-D distance between the car and train:

`D_(2D)^2(t) = C^2(t) + T^2(t)`

We can simply use C(t) and T(t) as the legs of the triangle based on how we defined our axes! Now, let us substitute the functions we found for C(t) and T(t):

`D_(2D)^2(t) = (25t-20)^2 + (-18t)^2`

Simplifying the squared terms:

`D_(2D)^2(t) = 625t^2 - 1000t + 400 + 324t^2 = 949t^2-1000t+400`

Now, let's take the square root of each side to isolate `D_(2D)(t)` :

`D_(2D)(t) = sqrt(949t^2-1000t+400)`

In order to minimize the distance found, we will now take the derivative of `D_(2D)(t)` and set it to zero (don't forget the chain rule!):

`(d(D_(2D)(t)))/dt = 1/2*(949t^2-1000t+400)^(-1/2) * (1898t - 1000) = 0`

Now, we can simplify to get rid of the 1/2 term in front:

`0=(949t - 500)/sqrt(949t^2-1000t+400)`

This looks daunting, but because we have "0" on the left, we can just multiply through by the square root to get rid of it!

`0 = 949t-500`

We can quickly solve this by adding 500 and dividing by 949 to give the following result:

`t = 500/949 = 0.527`

So, t = 0.527 gives us the time at which the minimum distance is achieved.

Let's first determine what our `D_(2D)(t)` value is at this time:

`D_(2D)(0.527) = sqrt(949(0.527)^2 - 1000(0.527) + 400)=11.7`

So, our minimum 2-D distance is 11.7 m.

Alright, now, we can go back and start considering the height of the overpass. You can convince yourself that the 3-D distance between your car and the train is simply the hypotenuse of another right triangle with its two legs being `D_(2D)` (11.7 m) and the height of the overpass (15 m). Again, we'll use the Pythagorean Theorem to solve for our overall distance.

`D_(3D)^2 =11.7^2 + 15^2`


`D_(3D) = sqrt(361.89) = 19.0 m`

And there's your answer. The minimum distance between you and the train is 19.0 m. I hope this helps!


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