# KINEMATICS Calculus using Distance, Velocity and Acceleration.

Position/Velocity/Acceleration Part 1: Definitions
Position/Velocity/Acceleration Part 1: Definitions

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KINEMATICS Calculus using Distance, Velocity and Acceleration

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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 cricket ball is hit and falls just short of getting a 6 The displacement, or its distance above the ground, is from the time the ball hits the bat until the ball falls onto the ground It can be defined in the following function where y is displacement in metres above the ground and t is the number of seconds since the ball was hit Distance is measured in metres y (t) = -14t 2 + 23t + 2

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Velocity y (t) = -14t 2 + 23t + 2 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) Velocity is in metres/sec

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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) Acceleration is in metres/sec/sec y (t) = -14t 2 + 23t + 2 y ’ (t) = -28t + 23 y ” (t) = -28

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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 metres, velocity in metres/sec, and acceleration in metres/sec/sec A change in direction can be indicated when a graph goes from negative to positive values or from positive to negative values

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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 metres. The function can be defined as f (t) = -30t 3 + 12t 2 + 8t +2 Find velocity and determine what velocity is at t = 1 Is the distance increasing or decreasing at t = 1 and explain Determine acceleration of the function and determine acceleration at t=1

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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 metres. The function can be defined as f (t) = -30t 3 + 12t 2 + 8t +2 Find velocity and determine what velocity is at t = 1 Is the distance increasing or decreasing at t = 1 and explain Determine acceleration of the function and determine acceleration at t=1

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Example Number 1 – Part A 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 metres. The function can be defined as f (t) = -30t 3 + 12t 2 + 8t +2 Find velocity and determine what velocity is at t = 1 Velocity is the derivative of acceleration so take the function’s derivative f (t) = -30t 3 + 12t 2 + 8t +2 f ’ (t) = -90t 2 + 24t + 8 Now since we know velocity we can plug into the equation for t. f’ (1)= -90(1) 2 + 24(1)+ 8 f’ (1)= -58metres/sec

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Example Number 1 – Part B 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 metres. The function can be defined as f (t) = -30t 3 + 12t 2 + 8t +2 Is the distance increasing or decreasing at t = 1 and explain? Now since we know velocity = -58 m/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.

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Example Number 1 – Part C 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 metres. The function can be defined as f (t) = -30t 3 + 12t 2 + 8t +2 Determine acceleration of the function and determine acceleration at t=1 We can find acceleration by just taking the derivative of velocity. f ’ (t) = -90t 2 + 24t + 8 f ” (t) = -180t + 24 Now we need to find acceleration at t = 1 f ” (1) = -180(1) + 24 f ” (1) = -156 m/sec/sec

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Example Number 2 Find the average velocity at t = 3 by using the average rate of change Hint: y2 – y1 x2 – x 1 010 24 417 621 t d (t)

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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

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Multiple Choice # 1 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

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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.

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Multiple Choice # 2 s (t) = t 2 – 20 Find the average velocity from t = 3 to t = 5 a.) 2b.)4c.)6d.)8

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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 – 3 2 2 Choice d.) 8

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Multiple Choice # 3 An objects distance from its starting point at time (t) is given by the equation d (t) = t 3 – 6t 2 – 4. What is the speed of the object when its acceleration is 0? a.) 2 b.) -24 c.) 22 d.) 44 e.) -12

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

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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

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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 = 3 +6 +62 2

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Hmmm? 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.

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Answer a.) v (1.5) = 1.5sin(1.5 2 ) = 1.167 Up, because v (1.5) is greater than 0 b.) a (t) = v’ (t) = sin t 2 + 2 t 2 cos t 2 a (1.5) = v’ (1.5) = -2.048 or -2.049 No; v is decreasing at 1.5 because v’ (1.5) is less than 0

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Continued Answer 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 = 7 2 2 y (t) = – 1cos t 2 + 7 2 2 y (2) = – 1cos4 + 7 = 3.826 or 3.827 2 2 d.) distance = the integral from 0 to 2 of the absolute value of v(t) = 1.173

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