Suppose the position of an object moving horizontally along a ...

Question

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.
On what intervals is the speed increasing?
f(t) = t² - 4t; 0 ≤ t ≤ 5

On what intervals is the speed increasing?

f(t) = t² - 4t; 0 ≤ t ≤ 5

Accepted Answer

Welcome back, everyone. In this problem, the position of a particle moving along a line is given by the function X of T equals T2 minus 8 T + 15, where X is in centimeters and the T is in seconds. Identify the intervals where the particle's speed is increasing. A says it's at T less than 5, B when T is greater than 8, C when T is less than 4, and D when T is greater than 4. Now how can we identify the intervals where the particle's speed is increasing? Well, notice here that we have our position function, OK? And we know that the velocity, the speed is the derivative of the position function, while the acceleration is the derivative of the velocity. So if we can find those derivatives, then we should be able to find the intervals where the particle's speed, the velocity is increasing. So first, let's identify the velocity function, OK? Now as we said, we know. That the velocity is the derivative. OK. The velocity of the derivative with respect to T of our position function. So the derivative with respect to T of X of T. So, that's going to be the derivative with respect to T, OK, of T2 minus 8 T + 15. I know if we apply the power rule here, then the derivative of T squared is 2T. The derivative of negative 8T is neg8, and the derivative of 15 is 0. So a velocity function is 2T minus 8. Next, let's find acceleration, the derivative of the velocity function, OK? So now we know that the acceleration function A of T is going to be the derivative with respect to T of V of T. That's equal to the derivative with respect to tea. Of 2 T minus 8. And now the derivative of 2 t by the pore rule equals 2, while the derivative of negative 8 equals 0. Now that we have our acceleration function, we can determine the intervals where the particle's speed is increasing. Now, it increases when the velocity and acceleration have the same sign, OK? And now, since the acceleration A of T is equal to 2. That means it's always positive, and thus we need to find. OK, we need to find when VFT is also going to be positive. OK. Now how do we figure that out? Well, we know that V of T equals 2 T minus 8. So if it's always positive, then 2 T minus 8 is going to be greater than 0. Solving this inequality, that means 2T is going to be greater than 8, and thus T is going to be greater than 8 divided by 2, which is 4. Therefore, the particle's speed is increasing for T greater than 4. If we look back at our answer twice, that means D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.
On what intervals is the speed increasing?
f(t) = t² - 4t; 0 ≤ t ≤ 5

Key Concepts

Position Function

Velocity and Speed

Acceleration

Watch next

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.On what intervals is the speed increasing?f(t) = t2 - 4t; 0 ≤ t ≤ 5

Key Concepts

Position Function

Velocity and Speed

Acceleration

Watch next

Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.
On what intervals is the speed increasing?
f(t) = t² - 4t; 0 ≤ t ≤ 5