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# 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 $t = 1$.

## A.$4.42,5.35,6.094,6.2614,6.27814$B.6.28 $\mathrm{m} / \mathrm{s}$

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PV

Pere V.

September 30, 2020

PV

Pere V.

September 30, 2020

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

this problem. Number six of Stewart eighth Edition, section 2.1 And the problem says if a rock is thrown upward on the planet Mars with a velocity of 10 m per second, it's heightened meters T seconds after it is done is given by y equals 10 T minus 1.86 He squared party. We find the average velocity over the given time intervals. So to review average velocity is to find very similarly to a slope where the rises change in wine and the run is changing team. And so if we look at these points for part a one comma two, one comma 1.5 one common 1.1111 point one and one Come at 1.1. We see that the time of interest is t equals one and that's also alluded, alluded to in part B. So in order to have the reference for finding average velocity. So we need a reference we need We need to know basically what is maybe the final y minus the initial y the final time, minus the initial time. So this will be important test. We're going to use this equation and find what is the value? Why first, Initially, when the time is one, we get 10 times one minus 1.86 times one squared, which is one. And this gives us eight. Wait one for And that is the location. Meters. This is the location of the Iraq after one second. So 88.14 m will be ry Initial time t equals one is going to be our time initial. And in this way we're going to use these points to calculate the slope which we identifying this problem as the average velocity. So we're going to refer to the spreadsheet to confirm our calculations for each of these points at the first time. One comma two, this is These are time intervals. So it's from time equals 12 time equals two. What we're going to do is we're going to calculate at time equals two. Use the equation to find the why at the time equals two. Then we're going to take the difference between the final y, which is this one subtracting the initial line, which we calculated at 8.14 The change in time is of course, to minus one. I don't mind this initial, and the average velocity is the ratio of the rise delta y over the run Delta T So for the first part of party, the average velocity is 4.42 for the interval, one 1.5, the average velocity turns out to be 5.35 for the next interval, 1 to 1.1. The average velocity 6.94 the next interval, 1 to 1.1. The average velocity 6.2614 And finally, for the interval 1 to 1.1 The average velocity is 6.27814 Now these average velocities are the answers to partying. To find out what part B will be, we need to understand that instantaneous velocity is something we can estimate if we continue this process even closer to T cold one. So I've continued the calculations two more times until we have a time that's very close to one, at which point we see that the average velocity is converging to approximately 6.28 As you can see, it gets closer and closer to 6.28 So our final answer or party is that the instantaneous velocity AT T equals one is six 0.28 meters per second.

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp