KINEMATICS

Calculus using Distance, Velocity and Acceleration

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) = -14t2 + 23t + 2

Velocity

y (t) = -14t2 + 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

Acceleration

Acceleration occurs when velocity changes

It is the derivative of velocity and the 2nd 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) = -14t2 + 23t + 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 decreasingIf velocity and acceleration have the same sign,

speed increases and vice versaNegative acceleration means velocity is decreasingDistance is measured in metres, velocity in

metres/sec, and acceleration in metres/sec/secA 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 metres.

The function can be defined as f (t) = -30t3 + 12t2 + 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

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) = -30t3 + 12t2 + 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

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) = -30t3 + 12t2 + 8t +2

a) Find velocity and determine what velocity is at t = 1Velocity is the derivative of acceleration so take the

function’s derivativef (t) = -30t3 + 12t2 + 8t +2f ’ (t) = -90t2 + 24t + 8Now since we know velocity we can pluginto the equation for t.f’ (1)= -90(1)2 + 24(1)+ 8 f’ (1)= -

58metres/sec

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) = -30t3 + 12t2 + 8t +2

b) 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.

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) = -30t3 + 12t2 + 8t

+2c) Determine acceleration of the function and

determine acceleration at t=1

We can find acceleration by just taking thederivative of velocity.f ’ (t) = -90t2 + 24t + 8f ” (t) = -180t + 24Now we need to find acceleration at t = 1f ” (1) = -180(1) + 24 f ” (1) = -156

m/sec/sec

Example Number 2

Find the averagevelocity at t = 3 byusing the average

rateof change

Hint: y2 – y1 x2 – x 1

0 10

2 4

4 17

6 21

t

d (t)

How To Solve It

To use this the formula we can pick anytwo y values and any two x values and findits slope or average rate of change. The bestestimate would be to use the values closest to t = 3

17 – 4 = 134 – 2 2

This answer is the average velocity at t = 3

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.) 2 b.) 4 c.) 6 d.) 7 e.) 8

Answer key to Multiple Choice # 1

Answer = c.) 6

The reason for this is that the graph goesor changes from positive to negative valuesat t = 6 indicating a change in direction.

Multiple Choice # 2

  s (t) = t2 - 20

Find the average velocity from t = 3 to t = 5

a.) 2 b.)4 c.)6 d.)8

Answer Key to Multiple Choice #2

For this problem we must use the averagevelocity formula

s(5) – s(3) = 5 – (-11) = 16 = 8 5 – 3 2 2

Choice d.) 8

Multiple Choice # 3

An objects distance from its starting pointat time (t) is given by the equation d (t) = t3 - 6t2 - 4. What is the speed ofthe 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 distancethen the derivative of velocity

d’ (t) = 3t2 - 12t d” (t) = 6t – 12

Now you must solve for t by setting d” oracceleration = 0

6t – 12 = 0 = 6t = 12 t = 2 +12 +12 6 6

Multiple Choice # 4

A particle moves along the x-axis so that itsposition at time t is given by d (t) = t2 – 6t + 5. For what value of t is thevelocity of the particle zero?

a.) 1 b.) 2 c.) 3 d.) 4 e.) 5

Answer Key to Multiple choice # 4

Choice a.) 3

First we find the derivative of the distancethan we set = 0 just like multiple choiceproblem # 3

V(t) = 2t – 6 2t - 6 = 0 2t = 6 t = 3

+6 +6 2 2

Hmmm?

A particle moves along the y-axis with the velocity given

by v (t) = t sin ( t2 ) 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

a.) v (1.5) = 1.5sin(1.52) = 1.167Up, because v (1.5) is greater than 0

b.) a (t) = v’ (t) = sin t2 + 2 t2 cos t2 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

Continued Answer

c.) y(t) = the integral of v(t) dt = the integral of tsin t2 dt = - cos t2 + C

2y (0) = 3 = -1+ C = 7 2 2

y (t) = - 1cos t2 + 7 2 2

y (2) = - 1cos4 + 7 = 3.826 or 3.827 2 2

d.) distance = the integral from 0 to 2 of theabsolute value of v(t) = 1.173