Analyzing Motion Using Graphs

Question

[Technology Exercise] Exercises 31–34 give the position function s = f(t) of an object moving along the s-axis as a function of time t. Graph f together with the velocity function v(t) = ds/dt = f'(t) and the acceleration function a(t) = d²s/dt² = f''(t). Comment on the object’s behavior in relation to the signs and values of v and a. Include in your commentary such topics as the following:

b. When does it move to the left (down) or to the right (up)?

s = 200t - 16t², 0 ≤ t ≤ 12.5 (a heavy object fired straight up from Earth’s surface at 200 ft/sec)

Accepted Answer

Welcome back, everyone. A rocket is launched vertically up and its height above the ground is given by the function H of T equals 400c minus 50 T2 where H is in meters and T is in seconds. The graph below shows H of T along with its velocity function V of T equals the H divided by DT, and the acceleration function A of T equals the second derivative of H with respect to time for the interval. Of 0 is less than or equal to T, less than or equal to 8 seconds, using the graph determined the time intervals when the rocket is ascending and when it is descending. So we have a two part problem. Let's begin with understanding when the rocket is ascending. It basically means that it is going up, and the condition for this is to have velocity. That is greater than 0 positive velocity. So we want to ensure that velocity is greater than 0. Were we given the velocity function, which is the H divided by DT. That's our velocity function, and it is given in green. So when is it positive? Well, whenever it is above the x axis, so we can see that in the given domain that's from 0 up to 4, right, because we have And X intercept or the T intercept in this case at T equals 4. So we can see that velocity is positive. If time is between 0 inclusive and 4 not inclusive, because 4 is our inner self. Now, when is a D setting, we're going to label it as D. Well, essentially the condition for going down. It is simply to have a negative velocity. So when the velocity is negative, our object is descending. So this means velocity must be negative, or basically our line must be below the T axis. It must be corresponding to the negative values. With respect to our vertical axis, so what can we tell? Well, essentially we have all of that region starting with 4 up to the end point of the domain at T equals 8, which is within the domain. So we can say that time for this case is between 4 and 8 inclusive. And those will be our answers so we can say that it is ascending. When velocity is positive and that's for time values between 0 inclusive and 4 not inclusive in seconds and descending. When time is greater than 4 seconds, but less than or equal to 8 seconds, and those will be our final answers to this problem. Thank you for watching.

Key Concepts

Position, Velocity, and Acceleration

Derivatives and Their Physical Interpretation

Sign of Velocity and Acceleration

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