If a rock is thrown upward on the planet Mars with a...

If a rock is thrown upward on the planet Mars with a velocity of $ 10 m/s $, its height in meters $ t $ seconds after is given by $ y = 10t - 1.86t^2 $. (a) Find the average velocity over the given time intervals: (i) $ [1, 2] $ (ii) $ [1, 1.5] $ (iii) $ [1, 1.1] $ (iv) $ [1, 1.01] $ (v) $ [1, 1.001] $ (b) Estimate the instantaneous velocity when $ t = 1 $.

If a rock is thrown upward on the planet Mars with a velocity of $ 10 m/s $, its height in meters $ t $ seconds after is given by $ y = 10t - 1.86t^2 $.

(a) Find the average velocity over the given time intervals:
(i) $ [1, 2] $ (ii) $ [1, 1.5] $
(iii) $ [1, 1.1] $ (iv) $ [1, 1.01] $
(v) $ [1, 1.001] $
(b) Estimate the instantaneous velocity when $ t = 1 $.

Transcript

00:01 All right.

00:03 We are here to solve this calculus problem about throwing a rock upwards on mars.

00:07 So we're looking at the question text here.

00:11 We've got a starting velocity of 10 meters per second.

00:14 We have our height in meters given by y equals 10 t minus 1 .86 t squared.

00:20 We're asked to find the average velocity over these intervals.

00:25 So in order to find the average velocity, if we remember, i'll use this notation.

00:31 Since it's fairly common in calculus textbooks, that's actually going to be given by this definition.

00:43 So this right here is a vector, and when we're talking about a change in a vector, in this case it's our position, we need to take whatever our final position is.

00:55 Again, this is our position vector.

00:57 So this is a vector pointing from the origin to our position.

01:01 Take that final, minus the initial position vector.

01:08 Now this is where if you're clever with setting up your coordinate systems, usually you will set one of these two vectors at the origin so that one of these two vectors is zero.

01:19 That's the great thing about these vectors.

01:21 You can choose your coordinate system and make it so that either x -final or x -initial is equal to zero.

01:27 So for this case, we're throwing a rock upwards.

01:29 Basically, if we find its initial height and its final height, find a difference and divide by the time it took to get there, then we can get our average velocities as we were asked.

01:40 So rather than doing this individually every time, i like to do repeated operations like this by coding them, i did some command line coding an octave to get these calculations.

01:55 So for part a, i'm going to copy these from this other tab.

02:00 We can see that our average velocities are given by this v average up at the top.

02:05 Ignore the code, the numbers are what matter.

02:08 We've got 4 .42, 5 .35, 6 .09.

02:12 So i'm just going to keep two decimal places on these because i don't think we need to be any more precise than that.

02:17 Also, if this is for an ap class, maybe i should say we should keep three decimal places, since i know that's at least the convention for ap physics when i took it.

02:26 So our average velocities, the shortest time interval, sorry, for the longest time interval, we actually get the lowest average velocity.

02:33 We had 4 .42 meters per second over that interval...

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