Suppose a stone is thrown vertically upward from the edge of a...

Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t² + 32t + 48.
On what intervals is the speed increasing?

On what intervals is the speed increasing?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says determine the impulse at which the speed of the ball is increasing if the ball is launched straight up from the top of a building with an initial velocity of 40 ft per second from a height of 60 ft above the ground. The height and feet of the ball above the ground, 10 seconds after it launch is given by H of T, which is equal to -16 T2 + 40T + 60. We're given 4 possible choices as our answers. For choice A, we have the open interval from 0 to 3.55. For choice B, we have the open interval from 0 to 1.25. For choice C, we have the open interval of 1.25 to 3.55. And for choice D, we have the open interval from 1.55 to 3.25. Now, as to determine the intervals at which the speed of the ball is increasing. Now, in order for the speed of the ball to be increasing, both the sign of our velocity as well as our acceleration must be the same. So either both of those need to be negative or both need to be positive. So, the first thing we need to do is calculate our velocity as a function of time. So, we're going to be looking for VFT. And recall that our velocity is equal to the time derivative of our position function. So this is going to be equal to the derivative with respect to T of H of T. Which is equal to the derivative with respect to T of the quantity of minus 16 squad. Plus 4DT. Plus 60. So taking this derivative, this is going to be equal to. For the first term, we're going to use our constant multiple and power rule, so we'll have minus 16, multiplied by the exponent, which is 2, multiplied by T raised to the power of 2 minus 1. Then we'll have plus, for the next term, we'll have the derivative of 40 T. I take the derivative of T with respect to T, that gives me 1, we'll have 40 multiplied by 1, which is just 40. And then we'll have plus the derivative with respect to T of 60. So here I'm taking the derivative of constant, so that's going to be 0. So simplifying this expression, this is going to be equal to minus 32, T + 40. So now what we want to do is Set this expression equal to 0, because that's going to give me the and solve it for T, because that's gonna give me the time at which our velocity is switching signs. So we'll set VFT. Equal to 0, which is going to be equal to minus 32 T plus 40, and now we're gonna solve this for T. So, I'm gonna add 32 T to both sides. We'll have 32T. equal to 40, and now define both sides by 32, we'll see that T is equal to 1.25 seconds. So now we need to look to see um if our velocity is positive or negative on both sides of this time value. So when T is less than 1.25, This means that the minus 32 T value is going to be less than 40, and so that means we're gonna get an overall positive value. So that means that VFT. It's going to be greater than 0. And for T greater than 1.25. Our first term, the minus 32 T, is going to actually be greater than 40, so that means that VFT is going to be negative, the VFT is going to be less than 0. So now, what we need to do is calculate our acceleration. So, our acceleration, which is AFT. That's going to be equal to, and if you recall for acceleration, that is the derivative with respect to time of our velocity function. So that's gonna be the derivative with respect to T of VFT. And this is going to be equal to the derivative with respect to T. Of the quantity of minus 32 T. Plus 40. And taking this derivative, this is going to be equal to, for the first term, we're gonna get -32. And for the second term, We're gonna have plus 0 since it is a constant. This means that our acceleration is just equal to -32 ft per second squared. So here our acceleration is negative. So that means we also need our velocity to be negative. That means that T is going to be greater than 1.25. And so our our speed is going to be increasing until the ball hits the ground. So that's gonna be the uh upper bound of our time interval. So what we need to do now is figure out the time at which the ball hits the ground. So, for this, we'll use our position function, HFT. And set it equal to 0, since that is the location of our ground. So we have 0 is equal to minus 16 T2d. Plus 40T. Plus 60. And here we have a quadratic equation, so we can solve this using our quadratic formula. So we call for the quadratic formula, we're gonna have T is equal to minus 40. Plus or minus the square root of the quantity of 40 squad, minus 4 multiplied by -16, multiplied by 60. In quantity, and that whole quantity is divided by. The quantity of 2 multiplied by -16. So we can simplify this expression, we'll have T is equal to. The negative signs will cancel out in the numerator denominator, so we'll have 40. Plus or minus the square root, and when evaluate the term inside the square root, fine, we get 5440, and then that whole quantity is divided by 32. So we can actually um simplify this even further, we'll see that T. It's equal to 40. Plus or minus, and when I um evaluate the square root of 5440 in my calculator, I get 73. 0.76 and the whole quantity is divided by 32. Now, here for our time, we want to take the positive value. And so we'll use the plus sign in front of the 73.76 because if you use a minus sign, I'll get a negative time, which is not what we're looking for. So we're looking for a time that is greater than 1.25. So evaluate this expression with the positive sign. We'll have T is equal to 3.55, and here we've just ran into two decimal places. So now, if we come back up and write our our time interval as um an interval with the interval notation, that's going to be the open interval from 1.25 to 3.55. So, this is the answer that we were looking for for this problem. And if we compare that answer to our answer choices, we see that the correct answer choice corresponds to answer choice C. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Velocity and Speed

Acceleration

Critical Points and Intervals

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