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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) after $ t $ seconds is given by $ H = 10t - 1.86t^2 $.
(a) Find the velocity of the rock after one second.(b) Find the velocity of the rock when $ t = a $.(c) When will the rock hit the surface?(d) With what velocity will the rock hit the surface?
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07:36
Daniel Jaimes
07:02
Dominique Jan Tan
Calculus 1 / AB
Chapter 2
Limits and Derivatives
Section 7
Derivatives and Rates of Change
Limits
Derivatives
David Base G.
October 27, 2020
That was not easy, glad this was able to help
Harvey Mudd College
University of Michigan - Ann Arbor
Idaho State University
Lectures
04:40
In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.
In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.
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If a rock is thrown upward…
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if a rock is thrown upward…
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If a rock is thrown vertic…
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Alright so here we are on mars we throw up An object with the starting velocity of 10 m/s and we observed that our height with time is 10 t minus 1.86 T squared. And we're going to find out a whole bunch of different features regarding this um this motion. Okay so part A. We just want to find V. F. T. And V. F. T. Is just the derivative of our position ah equation. So we will use the power rule and get 10 -3.72 T. Using power rules. So that's just doing power role on both parts. Um we need to also find v. of one by the way we have some units I believe. Yes our unit will be meters per second. Once I plug in the number. So if if we want to know the velocity at one second, one second after it was launched then we would just plug in one for the T. So times one and we end up with 6.28 meters per second. So the velocity one second later Is eight, sorry 6.28 m/s. Alright, let's see the next thing we need to do now we need to just in general what is the for any time A. So that will give us 10 -3.72 a. That's just plugging in a 14. Alright, that was pretty easy. Okay, Part C. Part C is now the rock is gonna hit the surface and we want to find the time it takes. So we basically want to find out when this um the rocks that's a rock was thrown. And so when the rock um Uh is back down to zero So that would be down to zero m. So We can solve this by pulling out A. T. that gives me 10 -1.86 T. T. Times 10 minus 1.8 62 B. Zero either T. Is zero. Well that's the starting time or 10 -1.860 equals zero. Which comes out to be T equals 10 over 1.86. And that is 5.376 seconds. So that's the time we want the time that the rock hits back to mars Alright one more part to go party and part D. We just want the velocity when the rock hits the ground. So we're going to go ahead and plug in our time. And into the velocity equation which was 10 -3.72 times that time. And when it's like that in To our calculator we get -10 m/s. So just like on earth if I launched from ground level when it finally the rock returns, this is without any air resistance. When the rock returns it returns with the same speed but in the opposite direction which is then downwards. So anyway hopefully that helps have a wonderful day
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