Calculus using Distance, Velocity and Acceleration By Tony Farah

Distance (Displacement)  Distance can be defined as the difference between point a and point b as a function of time  An example would be that a baseball is hit and falls just shy of homerun  The displacement, or its distance above the ground, is from the time the ball hits the bat until the ball falls into center field  It can be defined in the following function where y is displacement in feet above the ground and t is the number of seconds since the ball was hit  Distance is measured in feet y (t) = -14t t + 2

Velocity y ’ (t) = -28t + 23  Velocity can be defined as how fast an object is going and in what direction  Velocity can also be defined as an instantaneous rate of change or a derivative  In this case the velocity is how fast the ball is hit and in what direction it travels  Velocity = dy/dt or y ’ (t) (y prime of t)  Velocity is in ft/sec

Acceleration  Acceleration occurs when velocity changes  It is the derivative of velocity and the 2 nd derivative of displacement  Negative acceleration means velocity is decreasing  It can be defined as dv/dt or y ’’ (t) (y double prime of t)  Acceleration is in ft/sec/sec y (t) = -14t t + 2 y ’ (t) = -28t + 23 y ” (t) = -28

Key Information  To find velocity at a particular time, substitute for t in y (t)  Negative velocity means distance is decreasing  If velocity and acceleration have the same sign, speed increases and vice versa  Negative acceleration means velocity is decreasing  Distance is measured in ft, velocity in ft/sec, and acceleration in ft/sec/sec  A change in direction can be indicated when a graph goes from negative to positive values or from positive to negative values

Example Number 1 Jane and some friends are playing kickball after school. Jane kicks the ball so hard it flies up and over the fence. As it rises and falls its distance above the ground is a function of time since the kick measured in feet. The function can be defined as f (t) = -30t t 2 + 8t +2 a)Find velocity and determine what velocity is at t = 1 b)Is the distance increasing or decreasing at t = 1 and explain c)Determine acceleration of the function and determine acceleration at t=1

Part A Velocity is the derivative of acceleration so take the function’s derivative f (t) = -30t t 2 + 8t +2 f ’ (t) = -90t t + 8 Now since we know velocity we can plug into the equation for t. f’ (1)= -90(1) (1)+ 8 f’ (1)= -58 feet/sec

Part B Now since we know velocity = -58 ft/sec we can determine what this means in terms of distance. The velocity here is decreasing because the its sign is negative at t = 1. Whenever velocity is equals a negative number its distance is always decreasing. From this we can also concur that positive velocity means that distance is increasing.

Part C We can find acceleration by just taking the derivative of velocity. f ’ (t) = -90t t + 8 f ” (t) = -180t + 24 Now we need to find acceleration at t = 1 f ” (1) = -180(1) + 24 f ” (1) = -156 ft/sec/sec

Example Number 2 Find the average velocity at t = 3 by using the average rate of change Hint: y2 – y1 x2 – x 1 x2 – x t d (t)

How To Solve It To use this the formula we can pick any two y values and any two x values and find its slope or average rate of change. The best estimate would be to use the values closest to t = 3 17 – 4 = 13 4 – 2 2 This answer is the average velocity at t = 3

Multiple Choice # 1 (No Calculator) A bug begins to crawl up a vertical wire at time t = 0. The velocity v of the bug at time t, 0 < t < 8, is given by the function whose graph is shown above. At what value of t does the bug change direction? a.) 2b.) 4c.) 6d.) 7 e.) 8

Answer key to Multiple Choice # 1 Answer = c.) 6 The reason for this is that the graph goes or changes from positive to negative values at t = 6 indicating a change in direction.

Multiple Choice # 2 (with calculator) s (t) = t s (t) = t Find the average velocity from t = 3 to t = 5 a.) 2b.)4c.)6d.)8

Answer Key to Multiple Choice #2 For this problem we must use the average velocity formula s(5) – s(3) = 5 – (-11) = 16 = 8 5 – – Choice d.) 8

Multiple Choice # 3 (no calculator) An objects distance from its starting point at time (t) is given by the equation d (t) = t 3 - 6t What is the speed of the object when its acceleration is 0? a.) 2 b.) -24 c.) 22 d.) 44 e.) -12

Answer Key to Multiple Choice # 3 Choice a.) 2 You must take the derivative of distance then the derivative of velocity d’ (t) = 3t t d” (t) = 6t – 12 Now you must solve for t by setting d” or acceleration = 0 6t – 12 = 0 = 6t = 12 t =

Multiple Choice # 4 A particle moves along the x-axis so that its position at time t is given by d (t) = t 2 – 6t + 5. For what value of t is the velocity of the particle zero? a.) 1b.) 2c.) 3d.) 4e.) 5

Answer Key to Multiple choice # 4 Choice a.) 3 First we find the derivative of the distance than we set = 0 just like multiple choice problem # 3 V(t) = 2t – 62t - 6 = 02t = 6 t =

Free Response # 1 (you may use your calculator) Professor Frink never anticipated Truckasaurus getting loose, but then again, he never foresaw the obvious dangers in reanimating the corpse of his long-dead father. Frinkie's not one for long-term planning. Anyway, Homer Simpson lies directly in the path of the flame-spewing juggernaut, with only the meager acceleration of the family station wagon standing between him and utter destruction. Assume Homer's velocity (in feet per second) is given by the equation, where t is measured in seconds and. Answer the following questions based on the given information, accurate to the thousandths place. (a) Is Homer traveling backwards at any time during the first 7 seconds of his escape attempt? If so, during what time interval(s) is the gear shift in reverse? (b) What is Homer's average velocity from t = 2 to t = 5? (c) At what time(s) does Homer change from accelerating to decelerating, or vice versa?

Answer to Free Response # 1 Answer to Free Response # 1 (a) Homer is traveling backward whenever his velocity is negative. In other words, you're looking for the intervals of v(t) when its graph falls below the x-axis. If you graph v on your calculator, you'll find 2 x-intercepts in the interval [0,7]. It's advisable to use the calculator to find them: x = and x = Since the graph is below the x-axis between those x-intercepts, Homer's in reverse on the interval (.406, 4.182). (b) Since you're given the velocity function, and are asked to find its average value, you have to use the average value formula, which states that the average value of f on the interval [a,b] is. For this problem, a = 2 and b = 5. Since you're allowed to use a calculator, it would be a waste of time to compute the antiderivative by hand. The answer is (c) Homer will change from acceleration to deceleration (or vice versa) whenever the derivative of f has an x-intercept (since the derivative of the velocity function is the acceleration function). (Note that f also must cross the axis there, as well, not simply bounce off of it, or there will be no change in the sign of the graph.) The derivative of f is. The acceleration function has only one x-intercept on the interval [0,7]: x = seconds, at which Homer goes from decelerating to accelerating.

Free Response # 2 A particle moves along the y-axis with the velocity given by v (t) = t sin ( t 2 ) for t is greater than or equal to 0. a.) In which direction (up or down) is the particle moving at time t = 1.5? Why? b.) Find the acceleration of the particle at time t = 1.5. Is the velocity of the particle increasing at t = 1.5? Why or why not? c.) Given that y (t) is the position of the particle at time t and that y (0) = 3, find y (2). d.) Find the total distance traveled by the particle from t = 0 to t = 2.

Answer Key to Free Response # 2 a.) v (1.5) = 1.5sin(1.5 2 ) = Up, because v (1.5) is greater than 0 b.) a (t) = v’ (t) = sin t t 2 cos t 2 a (1.5) = v’ (1.5) = or No; v is decreasing at 1.5 because v’ (1.5) is less than 0

Continued Answer Key c.) y(t) = the integral of v(t) dt = the integral of tsin t 2 dt = - cos t 2 + C 2 y (0) = 3 = -1+ C = y (t) = - 1cos t y (2) = - 1cos4 + 7 = or d.) distance = the integral from 0 to 2 of the absolute value of v(t) = 1.173

Works Cited Mr. Watson Class Notes and Worksheets “Average velocity.” Chapter 20-4 Velocity and Acceleration. 7 June Foerster, Paul A. Calculus. EmeryVille: Steve Rasmussen, “Problem of the Year ” Calculus-Help. 7 June 2007.