Problem

The displacement (in meters) of a particle moving…

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Problem 14 Medium Difficulty

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?

Answer

a) Velocity at $t=1$ is 10-3.72=6.28 $\mathrm{m} / \mathrm{s}$
b) velocity of the rock when $\mathrm{t}=a$ is $10-3.72 \mathrm{a}$
c) The rock will hit the surface at $t=10/1.86=5.38$ seconds.
d) $10-3.72*5.38=-10 \mathrm{m} / \mathrm{s}$

October 27, 2020

That was not easy, glad this was able to help

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Video Transcript

good day. We are given with the height of the a certain object. In this case, rock from the surface of the planet. In this case, force as a function of time as a chick was 10 t -1.86 D squared where the HR were each year stands for the height from the surface in T. Time. All right. So the uh time here. So your height here is in meters and your time is in seconds. And we wish to find the velocity After 1 2nd. So that's at the equals one second. We also want to find the velocity at any time. T. So we say the 8th. Yeah. And we want to find the time when rock hits the ground. So you can imagine that rock here will return back to the ground. So that's the time that we wish to find in part C. And for partly we want to find the velocity upon hitting the surface. Now we note that. Mhm. The velocity, it's the time derivative of your position or in this case your height. So what we will do is to find the derivative of your height function with respect to time. And that will be the velocity function. So we have Again 10 T -1.8060 sq. So we will find the derivative of that. And we get V was 10- so 1.86. So, I'm going to applying chain rule here to obtain the velocity function in obtaining the derivative of the height function. Just and let's get get the derivative and that does get the velocity function. So I get this velocity function. So part A Find the time we find the velocity when T equals one. So I have 10 -3.72. Name swine. And that gives me 6.28. So not that the a positive value of 6.28 year indicates that the object or the rock in this case is moving upward. Mhm. For part B Let's assign any time eight. So we will simply substitute the with A. And so get 10 -3.72 E. Mhm. And then for part C we wish to find the time when the object hits the surface. So that's the time when your height becomes zero because again your height is measured with respect to the surface. So when Iraq is on the surface that's height equals zero. So what we will do is to set H20 and then we will solve for the so we can factor out first D. From here getting okay this expression and so we know that this is true If T is equal to zero why 10 -1.86 T equals zero. Or let's sell for D. Here. So I'm going to add 1.8060 on both equation side. So it will get Then it was 1.8060 and then I divide Both sides by 1.86. To solve for T equals So we get here to time values equals zero and equals five 38. Now, the significance of this T equals zero. Here is the initial time. Okay, Because at this time we assure certain that the object is really on the surface and then And then this time here is five point which is 5.8 is the time when the rock returns back to the surface. Mhm. Yeah. So for part D. Having known that it takes 5.38 seconds for the rock to return to the surface. So we will solve for the velocity at this time to get the velocity at which the rub hits the surface again. My velocity function is 10 -3.72 T. Then times 5.38. And so we get negative 10. I noticed that the negative value here means that the object is moving downward. Which just makes sense because during this time the object is no longer moving upward but is moving downward. All right, so, I hope all of these are clear. So, I'm just going to mark the our final answers for each of them. All right. So there you have it

Dominique Jan T.

Mindanao State University- Iligan Institute of Technology

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Problem

The displacement (in meters) of a particle moving…

Problem 14 Medium Difficulty

Answer

a) Velocity at $t=1$ is 10-3.72=6.28 $\mathrm{m} / \mathrm{s}$
b) velocity of the rock when $\mathrm{t}=a$ is $10-3.72 \mathrm{a}$
c) The rock will hit the surface at $t=10/1.86=5.38$ seconds.
d) $10-3.72*5.38=-10 \mathrm{m} / \mathrm{s}$

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a) Velocity at $t=1$ is 10-3.72=6.28 $\mathrm{m} / \mathrm{s}$b) velocity of the rock when $\mathrm{t}=a$ is $10-3.72 \mathrm{a}$c) The rock will hit the surface at $t=10/1.86=5.38$ seconds.d) $10-3.72*5.38=-10 \mathrm{m} / \mathrm{s}$

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Kristen K.

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Michael J.

Joseph L.

Precalculus Review - Intro

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Anna Marie V.

Caleb E.

Samuel H.

Joseph L.

a) Velocity at $t=1$ is 10-3.72=6.28 $\mathrm{m} / \mathrm{s}$
b) velocity of the rock when $\mathrm{t}=a$ is $10-3.72 \mathrm{a}$
c) The rock will hit the surface at $t=10/1.86=5.38$ seconds.
d) $10-3.72*5.38=-10 \mathrm{m} / \mathrm{s}$