Project

XII.) Kinematics of a Particle • Where are we?

1

Particle Kinematics

Rectilinear Motion

Curvilinear Motion

Single Coordinates

Rectangular Coordinates

Projectile Motion

Constant Acceleration?

Constant Acceleration Equations

Yes NO Differential Relationships

Normal and Tangential

Coordinates

Polar CoordinatesB. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

• Define a radial component, given by �𝑢𝑢𝑟𝑟, outward from some origin, 𝑂𝑂.

• The positive direction of the radial component is pointing outward from the origin, 𝑂𝑂, to the particle.

• Define the positive transverse component to be perpendicular to the radial component, measured by an angle, 𝜃𝜃, counterclockwise from a fixed reference, given by �𝑢𝑢𝜃𝜃.

• The coordinate system moves with the particle, but the origin is not necessarily at the particle.

2

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

• Note that the positive directions of the 𝑟𝑟 and 𝜃𝜃 coordinates are defined by the unit vectors, �𝑢𝑢𝑟𝑟, and �𝑢𝑢𝜃𝜃, as shown below.

3

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

• Keep in mind that the user chooses the origin. • Shown below is another valid coordinate system

for the particle at this instant.

4

𝑂𝑂

�𝑢𝑢𝑟𝑟

�𝑢𝑢𝜃𝜃 𝜃𝜃

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

– i.) Position • At any instant, the position of the particle is:

𝑟𝑟 = 𝑟𝑟�𝑢𝑢𝑟𝑟

– ii.) Velocity • We can differentiate the position to come up with

an expression for velocity: �⃗�𝑣 = 𝑑𝑑𝑟𝑟

𝑑𝑑𝑑𝑑

= 𝑑𝑑 𝑑𝑑𝑑𝑑

(𝑟𝑟�𝑢𝑢𝑟𝑟) = �̇�𝑟�𝑢𝑢𝑟𝑟 + 𝑟𝑟�̇𝑢𝑢𝑟𝑟

5

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

• We can use a similar procedure from before to find �̇𝑢𝑢𝑟𝑟, but this will be left as an exercise to perform on your own.

• Thus, without proof, we can say that: �⃗�𝑣 = �̇�𝑟�𝑢𝑢𝑟𝑟 + 𝑟𝑟�̇�𝜃�𝑢𝑢𝜃𝜃

• a.) Magnitude

𝑣𝑣 = �⃗�𝑣 = 𝑣𝑣𝑟𝑟2 + 𝑣𝑣𝜃𝜃 2

= �̇�𝑟2 + 𝑟𝑟�̇�𝜃 2

6

𝑣𝑣𝑟𝑟 𝑣𝑣𝜃𝜃

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

• b.) Direction – The direction is given in terms of �𝑢𝑢𝑟𝑟 and �𝑢𝑢𝜃𝜃.

�⃗�𝑣 = �̇�𝑟�𝑢𝑢𝑟𝑟 + 𝑟𝑟�̇�𝜃�𝑢𝑢𝜃𝜃

7

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

– iii.) Acceleration: • We can differentiate the velocity to come up with

the acceleration �⃗�𝑎 = �̈�𝑟�𝑢𝑢𝑟𝑟 + �̇�𝑟 �̇𝑢𝑢𝑟𝑟 + �̇�𝑟�̇�𝜃�𝑢𝑢𝜃𝜃 + 𝑟𝑟�̈�𝜃�𝑢𝑢𝜃𝜃 + 𝑟𝑟�̇�𝜃�̇𝑢𝑢𝜃𝜃

= �̈�𝑟�𝑢𝑢𝑟𝑟 + �̇�𝑟 �̇�𝜃�𝑢𝑢𝜃𝜃 + �̇�𝑟�̇�𝜃�𝑢𝑢𝜃𝜃 + 𝑟𝑟�̈�𝜃�𝑢𝑢𝜃𝜃 + 𝑟𝑟�̇�𝜃(−�̇�𝜃�𝑢𝑢𝑟𝑟) = �̈�𝑟 − 𝑟𝑟�̇�𝜃2 �𝑢𝑢𝑟𝑟 + 𝑟𝑟�̈�𝜃 + 2�̇�𝑟�̇�𝜃 �𝑢𝑢𝜃𝜃

• a). Magnitude

�⃗�𝑎 = 𝑎𝑎𝑟𝑟2 + 𝑎𝑎𝜃𝜃 2

= �̈�𝑟 − 𝑟𝑟�̇�𝜃 2

+ 𝑟𝑟�̈�𝜃 + 2�̇�𝑟�̇�𝜃 2

8

B. Slaboch 2017

XII.) Kinematics of a Particle • Coordinate System Summary

9

𝑟𝑟 = 𝑥𝑥 ̂𝚤𝚤 + 𝑦𝑦 ̂𝚥𝚥 + 𝑧𝑧�𝑘𝑘Position

Velocity

Acceleration

�⃗�𝑣 = �̇�𝑥 ̂𝚤𝚤 + �̇�𝑦 ̂𝚥𝚥 + �̇�𝑧�𝑘𝑘

�⃗�𝑎 = �̈�𝑥 ̂𝚤𝚤 + �̈�𝑦 ̂𝚥𝚥 + �̈�𝑧�𝑘𝑘

Rectangular Coordinates Polar Coordinates

𝑟𝑟 = 𝑟𝑟�𝑢𝑢𝑟𝑟

�⃗�𝑣 = �̇�𝑟�𝑢𝑢𝑟𝑟 + 𝑟𝑟�̇�𝜃�𝑢𝑢𝜃𝜃

�⃗�𝑎 = �̈�𝑟 − 𝑟𝑟�̇�𝜃2 �𝑢𝑢𝑟𝑟 + 𝑟𝑟�̈�𝜃 + 2�̇�𝑟�̇�𝜃 �𝑢𝑢𝜃𝜃

Normal and Tangential Coordinates

�⃗�𝑎 = �̇�𝑣�𝑢𝑢𝑑𝑑 + 𝑣𝑣2

𝜌𝜌 �𝑢𝑢𝑛𝑛

𝑟𝑟 = 0

�⃗�𝑣 = 𝑣𝑣�𝑢𝑢𝑑𝑑

B. Slaboch 2017

XII.) Kinematics of a Particle – Given all of these coordinate systems, how do you know

which one to use? • Look for cues in the problem

– Is the motion along a straight line path? » Use the differential relationships for rectilinear motion and a single

coordinate axis – Is the path curved?

» If the path is curved we probably want polar or normal and tangential coordinates

– Is the path even known? » We can’t use normal and tangential components without a known

path

10

B. Slaboch 2017

XII.) Kinematics of a Particle – Given all of these coordinate systems, how do you know

which one to use? • Look for cues in the problem

– Is there decoupling of the directions (i.e., projectile motion)? » Probably use a Cartesian coordinate system if there is

– Is the radial direction changing with time? » Polar coordinates would be a great option

• Then practice, practice, practice.

11

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

– EX. • A particle, 𝑃𝑃, moves in a flower pattern as shown

initially from 𝜃𝜃 = 0. If the flower pattern is described by:

𝑟𝑟 = 5 cos(2𝜃𝜃) and the angular velocity is:

�̇�𝜃 = 3𝑡𝑡2

determine the velocity and acceleration vector of 𝑃𝑃 when 𝑡𝑡 = 0.806 s and 𝜃𝜃 = 30∘.

12

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

– EX.

13

𝑟𝑟 𝜃𝜃

𝑟𝑟 = 5 cos 2𝜃𝜃

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

– EX. • We can start by defining the coordinate system:

14

𝑟𝑟 𝜃𝜃

𝑟𝑟 = 5 cos 2𝜃𝜃 �𝑢𝑢𝑟𝑟

�𝑢𝑢𝜃𝜃

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

– EX. • Given:

– Radial distance as a function of 𝜃𝜃 – Angular velocity as a function of time – Initial angle, 𝜃𝜃 0 = 𝜃𝜃0 = 0

• Find: – Velocity vector of 𝑃𝑃 when 𝑡𝑡 = 0.806 s and 𝜃𝜃 = 30∘

– Acceleration vector of 𝑃𝑃 when 𝑡𝑡 = 0.806 s and 𝜃𝜃 = 30∘

15

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

– EX. • Execution:

– We can start with the velocity and acceleration equations and determine which terms we know, and which terms we need to find:

– Which of these terms do we know directly, if any?

16

�⃗�𝑣 = �̇�𝑟�𝑢𝑢𝑟𝑟 + 𝑟𝑟�̇�𝜃�𝑢𝑢𝜃𝜃

�⃗�𝑎 = �̈�𝑟 − 𝑟𝑟�̇�𝜃2 �𝑢𝑢𝑟𝑟 + 𝑟𝑟�̈�𝜃 + 2�̇�𝑟�̇�𝜃 �𝑢𝑢𝜃𝜃

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

– EX. • Execution:

– We can quickly find �̇�𝜃 and �̈�𝜃

��̇�𝜃 = 3𝑡𝑡2 𝑑𝑑=0.806

= 1.95 rad

s

– Differentiating �̇�𝜃 gives:

��̈�𝜃 = 6𝑡𝑡 𝑑𝑑=0.806

= 4.836 rad s2

17

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

– EX. • Execution:

– Similarly, we can differentiate 𝑟𝑟 twice to find �̇�𝑟 and �̈�𝑟: �𝑟𝑟 = 5 cos 2𝜃𝜃 𝜃𝜃=30∘

= 2.5 m

��̇�𝑟 = −10 �̇�𝜃sin 2𝜃𝜃 𝜃𝜃=30∘ �̇�𝜃=1.95

= −16.88 m s

��̈�𝑟 = −20�̇�𝜃2 cos 2𝜃𝜃 − 10�̈�𝜃 sin 2𝜃𝜃 𝜃𝜃=30∘ �̇�𝜃=1.95 �̈�𝜃=4.836

= −80 m s2

18

B. Slaboch 2017

XII.) Kinematics of a Particle • G.) Polar Coordinates

– EX. • Thus, substituting in numbers in the velocity and acceleration vectors gives:

�⃗�𝑣 = �̇�𝑟�𝑢𝑢𝑟𝑟 + 𝑟𝑟�̇�𝜃�𝑢𝑢𝜃𝜃 = −16.88�𝑢𝑢𝑟𝑟 + 2.5 1.95 �𝑢𝑢𝜃𝜃 = −16.88�𝑢𝑢𝑟𝑟 + 4.87�𝑢𝑢𝜃𝜃

m s

�⃗�𝑎 = �̈�𝑟 − 𝑟𝑟�̇�𝜃2 �𝑢𝑢𝑟𝑟 + 𝑟𝑟�̈�𝜃 + 2�̇�𝑟�̇�𝜃 �𝑢𝑢𝜃𝜃 = −80 − 2.5 1.95 2 �𝑢𝑢𝑟𝑟 + 2.5 4.836 + 2 −16.88 1.95 �𝑢𝑢𝜃𝜃 = [89.5�𝑢𝑢𝑟𝑟 − 53.7�𝑢𝑢𝜃𝜃]

m s2

19

B. Slaboch 2017

XII.) Kinematics of a Particle • H.) Cylindrical Coordinates

• If particles move in 3D space, we can use cylindrical coordinates.

• We can add a 𝑧𝑧 component to the 𝑟𝑟 and 𝜃𝜃 we already have from polar coordinates

20

B. Slaboch 2017

XII.) Kinematics of a Particle • H.) Cylindrical Coordinates

• Since 𝑧𝑧 is rectangular (and fixed), it follows the same expression we had previously, and we add it to our polar coordinates as shown:

𝑟𝑟 = 𝑟𝑟�𝑢𝑢𝑟𝑟 + 𝑧𝑧�𝑢𝑢𝑧𝑧 �⃗�𝑣 = �̇�𝑟�𝑢𝑢𝑟𝑟 + 𝑟𝑟�̇�𝜃�𝑢𝑢𝜃𝜃 + �̇�𝑧�𝑢𝑢𝑧𝑧 �⃗�𝑎 = �̈�𝑟 − 𝑟𝑟�̇�𝜃2 �𝑢𝑢𝑟𝑟 + 𝑟𝑟�̈�𝜃 + 2�̇�𝑟�̇�𝜃 �𝑢𝑢𝜃𝜃 + �̈�𝑧�𝑢𝑢𝑧𝑧

21

B. Slaboch 2017

XII.) Kinematics of a Particle • H.) Cylindrical Coordinates

• Since 𝑧𝑧 is rectangular (and fixed), it follows the same expression we had previously, and we add it to our polar coordinates as shown:

𝑟𝑟 = 𝑟𝑟�𝑢𝑢𝑟𝑟 + 𝑧𝑧�𝑢𝑢𝑧𝑧 �⃗�𝑣 = �̇�𝑟�𝑢𝑢𝑟𝑟 + 𝑟𝑟�̇�𝜃�𝑢𝑢𝜃𝜃 + �̇�𝑧�𝑢𝑢𝑧𝑧 �⃗�𝑎 = �̈�𝑟 − 𝑟𝑟�̇�𝜃2 �𝑢𝑢𝑟𝑟 + 𝑟𝑟�̈�𝜃 + 2�̇�𝑟�̇�𝜃 �𝑢𝑢𝜃𝜃 + �̈�𝑧�𝑢𝑢𝑧𝑧

22

Add these terms to move from polar to cylindrical coordinates

B. Slaboch 2017

FE Exam Style Problems An object weighs 2 pounds and is travelling counterclockwise in a circular path of radius 5 ft. at a constant speed of 10 feet per second. The acceleration of the object at the instant shown is:

a.) �⃗�𝑎 = −40.87 ̂𝚤𝚤 − 6.28 ̂𝚥𝚥 ft s2

b.) �⃗�𝑎 = −18.13 ̂𝚤𝚤 − 8.45 ̂𝚥𝚥 ft s2

c.) �⃗�𝑎 = 86.63 ̂𝚤𝚤 − 5.81 ̂𝚥𝚥 ft s2

d.) �⃗�𝑎 = 74.91 ̂𝚤𝚤 − 4.22 ̂𝚥𝚥 ft s2

e.) �⃗�𝑎 = −45.67 ̂𝚤𝚤 − 32.84 ̂𝚥𝚥 ft s2

𝑥𝑥

𝑦𝑦

𝑟𝑟

25∘

FE Exam Style Problems An object weighs 2 pounds and is travelling counterclockwise in a circular path of radius 5 ft. at a constant speed of 10 feet per second. The acceleration of the object at the instant shown is:

a.) �⃗�𝑎 = −40.87 ̂𝚤𝚤 − 6.28 ̂𝚥𝚥 ft s2

b.) 𝒂𝒂 = −𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏�̂�𝒊 − 𝟏𝟏. 𝟒𝟒𝟒𝟒 ̂𝒋𝒋 𝐟𝐟𝐟𝐟 𝐬𝐬𝟐𝟐

c.) �⃗�𝑎 = 86.63 ̂𝚤𝚤 − 5.81 ̂𝚥𝚥 ft s2

d.) �⃗�𝑎 = 74.91 ̂𝚤𝚤 − 4.22 ̂𝚥𝚥 ft s2

e.) �⃗�𝑎 = −45.67 ̂𝚤𝚤 − 32.84 ̂𝚥𝚥 ft s2

𝑥𝑥

𝑦𝑦

𝑟𝑟

25∘

Normal-Tangential and Cartesian �𝑢𝑢𝑑𝑑

�𝑢𝑢𝑛𝑛 �𝑢𝑢𝑛𝑛 = − cos 25∘ ̂𝚤𝚤 − sin 25∘ ̂𝚥𝚥

�⃗�𝑎 = 20�𝑢𝑢𝑛𝑛

�⃗�𝑎 = 20 − cos 25∘ ̂𝚤𝚤 − sin 25∘ ̂𝚥𝚥

�⃗�𝑎 = 20 − cos 25∘ ̂𝚤𝚤 − sin 25∘ ̂𝚥𝚥

�⃗�𝑎 = −18.13 ̂𝚤𝚤 − 8.45 ̂𝚥𝚥 ft s2

Normal-Tangential

�⃗�𝑎 = �̇�𝑣�𝑢𝑢𝑑𝑑 + 𝑣𝑣2

𝜌𝜌 �𝑢𝑢𝑛𝑛

�⃗�𝑎 = 0�𝑢𝑢𝑑𝑑 + 102

5 �𝑢𝑢𝑛𝑛

�⃗�𝑎 = 20�𝑢𝑢𝑛𝑛

FE Exam Style Problems An object weighs 2 pounds and is travelling counterclockwise in a circular path of radius 5 ft. at a constant speed of 10 feet per second. The acceleration of the object at the instant shown is:

a.) �⃗�𝑎 = −40.87 ̂𝚤𝚤 − 6.28 ̂𝚥𝚥 ft s2

b.) �⃗�𝑎 = −18.13 ̂𝚤𝚤 − 8.45 ̂𝚥𝚥 ft s2

c.) �⃗�𝑎 = 86.63 ̂𝚤𝚤 − 5.81 ̂𝚥𝚥 ft s2

d.) �⃗�𝑎 = 74.91 ̂𝚤𝚤 − 4.22 ̂𝚥𝚥 ft s2

e.) �⃗�𝑎 = −45.67 ̂𝚤𝚤 − 32.84 ̂𝚥𝚥 ft s2

𝑥𝑥

𝑦𝑦

𝑟𝑟

25∘

FE Exam Style Problems An object weighs 2 pounds and is travelling counterclockwise in a circular path of radius 5 ft. at a constant speed of 10 feet per second. The acceleration of the object at the instant shown is:

a.) �⃗�𝑎 = −40.87 ̂𝚤𝚤 − 6.28 ̂𝚥𝚥 ft s2

b.) 𝒂𝒂 = −𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏�̂�𝒊 − 𝟏𝟏. 𝟒𝟒𝟒𝟒 ̂𝒋𝒋 𝒇𝒇𝒇𝒇 𝒔𝒔𝟐𝟐

c.) �⃗�𝑎 = 86.63 ̂𝚤𝚤 − 5.81 ̂𝚥𝚥 ft s2

d.) �⃗�𝑎 = 74.91 ̂𝚤𝚤 − 4.22 ̂𝚥𝚥 ft s2

e.) �⃗�𝑎 = −45.67 ̂𝚤𝚤 − 32.84 ̂𝚥𝚥 ft s2

𝑥𝑥

𝑦𝑦

𝑟𝑟

25∘

�𝑢𝑢𝜃𝜃 �𝑢𝑢𝑟𝑟 Polar

�⃗�𝑎 = �̈�𝑟 − 𝑟𝑟�̇�𝜃2 �𝑢𝑢𝑟𝑟 + 𝑟𝑟�̈�𝜃 + 2�̇�𝑟�̇�𝜃 �𝑢𝑢𝜃𝜃

�⃗�𝑎 = 0 − 5(2)2 �𝑢𝑢𝑟𝑟 + 5(0) + 2(0)(2) �𝑢𝑢𝜃𝜃 𝑣𝑣 = 𝑟𝑟�̇�𝜃

10 = 5�̇�𝜃

�̇�𝜃 = 2 rad

s

�̈�𝜃 = 0

�⃗�𝑎 = −20�𝑢𝑢𝑟𝑟

�𝑢𝑢𝑟𝑟 = cos 25∘ ̂𝚤𝚤 + sin 25∘ ̂𝚥𝚥

�⃗�𝑎 = −20(cos 25∘ ̂𝚤𝚤 + sin 25∘ ̂𝚥𝚥)

�⃗�𝑣 = �̇�𝑟�𝑢𝑢𝑟𝑟 + 𝑟𝑟�̇�𝜃�𝑢𝑢𝜃𝜃 �⃗�𝑣 = (0)�𝑢𝑢𝑟𝑟 + 𝑟𝑟�̇�𝜃�𝑢𝑢𝜃𝜃

�⃗�𝑎 = −18.13 ̂𝚤𝚤 − 8.45 ̂𝚥𝚥 ft s2

FE Exam Style Problems If a particle moves along a path such that 𝑟𝑟 = (2 cos 𝑡𝑡) ft. and 𝜃𝜃 = 𝑑𝑑

2 rad,

where 𝑡𝑡 is in seconds, determine the particles transverse component of acceleration.

a.) �⃗�𝑎𝜃𝜃 = − 5 2

cos 𝑡𝑡 b.) �⃗�𝑎𝜃𝜃 = −2 cos 𝑡𝑡 c.) �⃗�𝑎𝜃𝜃 = −2 sin 𝑡𝑡 d.) �⃗�𝑎𝜃𝜃 = −2 sin 𝑡𝑡 cos 𝑡𝑡 e.) �⃗�𝑎𝜃𝜃 = cos 𝑡𝑡

FE Exam Style Problems If a particle moves along a path such that 𝑟𝑟 = (2 cos 𝑡𝑡) ft. and 𝜃𝜃 = 𝑑𝑑

2 rad,

where 𝑡𝑡 is in seconds, determine the particles transverse component of acceleration.

a.) �⃗�𝑎𝜃𝜃 = − 5 2

cos 𝑡𝑡 b.) �⃗�𝑎𝜃𝜃 = −2 cos 𝑡𝑡 c.) 𝒂𝒂𝜽𝜽 = −𝟐𝟐𝐬𝐬𝐬𝐬𝐬𝐬𝒇𝒇 d.) �⃗�𝑎𝜃𝜃 = −2 sin 𝑡𝑡 cos 𝑡𝑡 e.) �⃗�𝑎𝜃𝜃 = cos 𝑡𝑡

�⃗�𝑎𝜃𝜃 = 𝑟𝑟�̈�𝜃 + 2�̇�𝑟�̇�𝜃

Polar

�⃗�𝑎𝜃𝜃 = 2 cos 𝑡𝑡 0 − 2(2 sin 𝑡𝑡) 1 2

�⃗�𝑎𝜃𝜃 = −2 sin 𝑡𝑡

FE Exam Style Problems The acceleration of a particle along a straight line is defined by 𝑎𝑎 = 2𝑡𝑡 − 9 m

s2 . At 𝑡𝑡 =

0, 𝑠𝑠 = 1 m and 𝑣𝑣 = 10 m s

. When 𝑡𝑡 = 9 s, determine the particle’s position.

a.) 𝑠𝑠 = −45.2 m b.) 𝑠𝑠 = 26.2 m c.) 𝑠𝑠 = −39.8 m d.) 𝑠𝑠 = 28.5 m e.) 𝑠𝑠 = −30.5 m

FE Exam Style Problems The acceleration of a particle along a straight line is defined by 𝑎𝑎 = 2𝑡𝑡 − 9 m

s2 . At 𝑡𝑡 =

0, 𝑠𝑠 = 1 m and 𝑣𝑣 = 10 m s

. When 𝑡𝑡 = 9 s, determine the particle’s position.

a.) 𝑠𝑠 = −45.2 m b.) 𝑠𝑠 = 26.2 m c.) 𝑠𝑠 = −39.8 m d.) 𝑠𝑠 = 28.5 m e.) 𝒔𝒔 = −𝟏𝟏𝟑𝟑. 𝟒𝟒 𝐦𝐦

𝑎𝑎 = 𝑑𝑑𝑣𝑣 𝑑𝑑𝑡𝑡

= 2𝑡𝑡 − 9

𝑎𝑎 = � 10

𝑣𝑣

𝑑𝑑𝑣𝑣 = � 0

𝑑𝑑

2𝑡𝑡 − 9 𝑑𝑑𝑡𝑡

𝑣𝑣 − 10 = 𝑡𝑡2 − 9𝑡𝑡

𝑣𝑣 = 𝑡𝑡2 − 9𝑡𝑡 + 10

𝑣𝑣 = 𝑑𝑑𝑠𝑠 𝑑𝑑𝑡𝑡

= 𝑡𝑡2 − 9𝑡𝑡 + 10

� 1

𝑠𝑠

𝑑𝑑𝑠𝑠 = � 0

𝑑𝑑

𝑡𝑡2 − 9𝑡𝑡 + 10 𝑑𝑑𝑡𝑡

𝑠𝑠 − 1 = 𝑡𝑡3

3 −

9𝑡𝑡2

2 + 10𝑡𝑡

𝑠𝑠 = 𝑡𝑡3

3 −

9𝑡𝑡2

2 + 10𝑡𝑡 + 1

𝑠𝑠 = 93

3 −

9 9 2

2 + 10 9 + 1 = −30.5 m