B) A certain coyote drops an anvil from a height of 31 m....

b) A certain coyote drops an anvil from a height of 31 m. After \(t\) seconds the anvils height is \(31 - 4.8t^2\) m. i. What is the anvil's velocity, speed and acceleration at time \(t\)? (Note: speed is the absolute value of velocity, and acceleration is the second derivative of the position function, or, equivalently, the first derivative of the velocity function) ii. How long will it take the anvil to hit the ground? iii. What would be the anvil's velocity at the moment of impact?

Transcript

00:01 A stone is thrown downward with initial velocity v -0 is equal to 8 meters per second.

00:11 And gravity, well, i'll write this, is acceleration, is also acting downward at magnitude 9 .8 meters per second squared.

00:24 And we need to calculate how much time it takes for the stone to reach the ground, which is a distance delta y of 4 .5.

00:33 400 meters away.

00:37 Acceleration is equal to the velocity, the derivative of the velocity with respect to time.

00:46 So we can get the velocity an equation for the velocity by integrating acceleration over time.

00:54 And note that our acceleration is constant.

00:59 So what we get is just that the velocity with respect to time is equal to 9 .8 meters per second squared times time plus some constant, which i'll call v0.

01:22 I call the constant v0 because when we plug in a t equal 0, we get v of t equals 0.

01:32 Equals and this term which has a factor of t is just zero and we get v not okay the initial speed so i'm just going to write this back in terms of variables we have a t plus v not and the position y as a function of t is equal to well first let's look at what speed is or velocity velocity is equal to the derivative of position with respect to time.

02:10 So that means that the position is the integral of the velocity over time.

02:17 So this is equal to integral of a t plus v not over time.

02:24 And we know that a and v not are constants...

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