Two cars are speeding towards an intersection from different...

Name: Two cars are speeding towards an intersection from different directions. Their accelerators are stuck and their brakes don't work. Car A is 300m south of the intersection and is travelling north at 15m/s. Car B is 500m west of the intersection and is travelling east at 20 m/s. Determine the closest distance the cars get to each other. (5 Marks)
Uploaded: 2022-07-02T16:47:56-08:00
Duration: 11 min 14 s
Channel: Joel Mueller
Description: Two cars are speeding towards an intersection from different directions. Their accelerators are stuck and their brakes don't work. Car A is 300m south of the intersection and is travelling north at 15m/s. Car B is 500m west of the intersection and is travelling east at 20 m/s. Determine the closest distance the cars get to each other. (5 Marks)

Two cars are speeding towards an intersection from different directions. Their accelerators are stuck and their brakes don't work. Car A is 300m south of the intersection and is travelling north at 15m/s. Car B is 500m west of the intersection and is travelling east at 20 m/s. Determine the closest distance the cars get to each other. (5 Marks)

Transcript

00:04 First thing i did was i set up a coordinate plane and let's call the intersection 0 .0.

00:24 And so what that means is that the car 1 is 300 south of the intersection to start with.

00:41 And it moves north at a rate of 15 meters per second.

00:48 So i'm going to add a 15t so that as time progresses, it'll get closer to the intersection.

00:55 Specifically, the point, let me rewrite this as a point.

01:01 You can describe the point with an equation.

01:05 You can describe it.

01:05 It's always going to be on the y -axis, so the x -corident, it's always going to be on the y -axis, so the x -cordinate is always going to be zero.

01:23 But the y -cordinate is going to be negative 300 plus 15 -t.

01:35 And the other car is always going to be on the x -axis.

01:42 So you can describe it as a point using the formula negative 500 plus 20t, comma zero.

01:59 And so we're trying to minimize distance.

02:01 So we're going to use the distance formula d.

02:05 And whenever i minimize distance, i always minimize d squared because it ends up being the same point either way.

02:13 And this way it gets rid of the square root.

02:16 I'm just taking the distance formula and i'm squaring it.

02:19 And so then i'll get the first point, negative 500 plus 20t squared.

02:34 Well, technically, okay, let me remind you what the distance formula looks like.

02:38 It's square root of x2 minus x1 squared plus y2 minus y1 squared.

02:50 And i'm squaring both sides to get rid of the square root.

02:53 Because when i minimize distance squared, it ends up being the same point as if i minimize distance.

03:02 So this is technically, it's technically what i'm doing is taking negative 500 plus 20t minus the other x coordinate and then squared.

03:21 Then i'm plus zero.

03:29 Let me do that in green.

03:38 Minus 300, negative 300 plus 15t, squared.

03:54 So let me write it neater more neatly.

04:01 It looks like negative 500 plus 20 t squared plus 300 minus 15 t squared.

04:21 And now i'm going to minimize this equation and set it equal to zero so i can minimize the distance.

04:26 So the derivative of d squared with respect to t is two.

04:34 I'm negative 500 plus 20t times 20 plus 2 times 300 minus 15 t times negative 15 and i'm going to set that equal to 0.

05:12 And i'm going to simplify a little bit...

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