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 19.6 m/s from a height of 24.5 m above the ground. The height (in meters) of the stone above the ground t seconds after it is thrown is s(t) = -4.9t² + 19.6t + 24.5.
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 a rocket is fired vertically upwards from the surface of the Earth with an initial velocity of 30 m per second. The height in meters of the rocket above the ground 10 seconds after it is fired is modeled by the function H of T is equal to -5 T2 + 30T + 35. Determine the intervals on which the speed of the rocket is increasing. We give them 4 possible choices as our answers. For choice A, we have the open interval from 0 to 3. For choice B, we have the open interval from 3 to 7. For choice C, we have the open interval from 0 to 7, and for choice D, we have the open interval from 2 to 7. Now we're asked to determine the interval on which the speed of the rocket is increasing. So we call for the speed of the rocket to be increasing, both the velocity and acceleration need to have the same sign. They either need to both be positive or both be negative. So we need to first calculate our velocity as a function of time. So we're going to be looking for VFT. And recall that our VAT 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 HFT. And then it's going to be equal to the derivative with respect to T. Of the quantity of minus 5 is squad. Plus 30T. Plus 35. They're taking a derivative of this um expression. For the first time, we're going to use our constant multiple and power rule. We will have minus 5, multiplied by the exponent, which is 2, multiplied by T raised to the 2 minus 1. Then we'll have plus for the derivative of the second term, which is 30T. The 30 is a constant, so we can move through the derivative, and I'll just have the derivative of T with respect to T, which is 1, so 30 multiplied by 1 is going to give me 30. For the second term, and then we'll have plus the derivative of 35 with respect to T. Here we're taking a derivative of a constant, and anytime we take a derivative of a constant, its derivative is going to be equal to 0, so we'll have plus 0 as the last term. So simplifying this expression, this is going to be equal to minus 10 T plus 30. So now we need to determine the points in which the sign of our velocity changes. So, to do that, we're gonna take our function here, VFT. And set it equal to 0. That means that 0 is going to be equal to minus 10, T plus 30. So solving this expression for T, we'll add 10 T to both sides of this equation. So we'll have 10 T is equal to 30. And dividing both sides of the equation by 10, we'll get T is equal to 3 seconds. So, we'll look at the case when T is less than 3. And In that case, We see that 10T is always going to be less than 30, so minus 10T plus 30 is always going to be a positive number. That means that VFT. is always going to be greater than 0. Now, if we look at the case when T is greater than 3, In that case, 10T is always going to be greater than 30, and so, minus 10T plus 30 is always going to be a negative number. That means that VFT is always going to be less than 0. So now I need to calculate my acceleration in order to compare the sign of the acceleration to the sign of the velocity. So, we're going to be looking for acceleration function, our acceleration as a function of time. So we'll be looking for AFT, and we call that our acceleration is equal to the time derivative of our velocity function. So AFT is going to be equal to the derivative with respect to T. Of VFT. That's gonna be equal to the derivative with respect to T. Of the quantity of minus 10 T. Plus 30. So taking a derivative of this expression, this is going to be equal to. For the first term, the minus 10 can move through the derivative, and then we'll just have the derivative T with respect to T. So, that is equal to 1, so we'll have minus 10 multiplied by 1, so that gives us minus 10 for the first term. The one we'll have plus the derivative of 30 with respect to T, and again, here we're taking the derivative of a constant, so its derivative is equal to 0. So this means that our acceleration is equal to -10. So here our acceleration is a constant value, and it is negative. So, since we need our veloci to also be negative in order for the speed to be increasing, we're gonna be looking for time intervals where T is greater than 3. So, Ter than 3 means that our rocket is in the air, and at some point it's going to hit the ground. So, T T3 is going to be the lower bound of our interval, and now we need to figure out the upper bound of our interval, which is when the rocket hits the ground. So, to do that, we're going to take our position function, HFT. And set it equal to 0, since that is the location of the ground. And this is going to be equal to. -5 T squared plus 30 T. Plus 35. And now we need to solve this equation for T, and here we have a quadratic equation, and we can actually factor our quadratic function here. So this is going to be equal to the quantity of minus 5 T. -5 in quantity, multiplying the quantity of T minus 7. So now what we want to do is set each of these factors equal to 0 and solve for T. So we'll have minus 5 T. -5 is equal to 0 for the first factor, and solving this for T we'll get T is equal to -1. For the second factor, we will have T minus 7 is equal to 0, and solving this for T, we'll get T is equal to 7. Here we're interested in our time value T that is greater than 3, so we're going to ignore the T equal to -1 solution. And so we're just going to be using our T equal to 7 as our upper bound or our interval. So that means that our time has to be between 3 and 7 seconds. So writing that in interval notation, we will have the open interval from 3 to 7. So that is the answer that Uh, the question was looking for. And so when we compare our answer here to our answer choices, we see that the correct answer choice is answer choice B. 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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