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KINEMATICS REVIEW VECTOR ALGEBRA - SUMMARY

▪ Magnitude – A numerical value with appropriate units.▪ Scalar is a quantity that is completely specified by magnitude. ▪ Vector requires both, magnitude and direction for a complete description. The main difference between scalars and vectors is difference between scalar and vector algebra – geometrical description of vector quantity:▪ the direction of a vector is the counterclockwise angle of rotation which that vector makes with positive direction of x - axis.

▪ Two vectors are equal if they have the same magnitude and directions. ▪ Vectors that have the same magnitude and the same direction are the same.

▪ Multiplying a vector by a scalar: Multiplying vector by 2 increases its magnitude by a factor 2. ▪ Opposite vectors: A⃗∧−A⃗

Multiplying by –½ changes the magnitude ½ times and reverses the direction

▪ Components of a Vector (x or horizontal component and y or vertical component) Any vector can be “resolved” into two components.

If you know the magnitude A and direction θ of a vector = (A, θ) you can find x and y components of that vector:

v⃗ = 34 m/s @ 48° . Find vx and vy

vx = (34 m/s )cos 48 ) = 23 m/s; vy = (34 m/s) sin 48° = 25 m/s

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If you know x and y components of a vector A

you can find the magnitude A and direction θ of that vector:

▪ Vector addition graphically – comparison between “head-to-tail” and “parallelogram” method Two methods for vector addition are equivalent. examples: Find F⃗1+ F⃗2=F⃗ using both methods.

The sum is the vector sum of the two individual vectors, known as the "resultant" or “net vector

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SUBTRACTION is adding opposite vector.

▪ ANALYTICALLY/NUMERICALLY:

1. step: Find components of the resultant vector

C⃗= A⃗+ B⃗ Cx=Ax+B x=A cosθA+BcosθB

C y=A y+B y=A sin θA+BsinθB

Fx = 4 N and Fy = 3 N .

Find magnitude (always positive!) and direction.

; = arc tan (¾) = 370if the vector is in the first quadrant;if not you find it from the picture.

examples: resultant velocity, resultant force or net force (the one that can replace all forces that are applied at the same time)

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2. step: Knowing components of the resultant vector, find its magnitude and direction:

C=¿ √C x2+C y

2 ; from the picture

C⃗=C (units ) , θ

EXAMPLE: F⃗1 = 68 N @ 24° F⃗2 = 32 N @ 65° Find

F⃗=F⃗1+ F⃗2

Fx = F1x + F2x = 68 cos240 + 32 cos650 = 75.6 N

Fy = F1y + F2y = 68 sin240 + 32 sin650 = 56.7 N

DEFINITIONS and FORMULAS for 1-D KINEMATICS

► Displacement – A change of position in particular direction. A distance in a given direction. Vector. Unit: meter (m)

► Average Velocity = displacementelapsed time

v⃗avg = xt

► Average Speed = dis tance traveledelapsed time vavg =

dt

► (Instantaneous) Velocity – 1. Value of velocity at a particular time. ► (Instantaneous) Speed – 1. Value of speed at a particular time.

► Acceleration

F=√Fx2+F y

2=94 .5N

= arc tan (56.7/75.6) = 36.90

a=∆v∆ t

= change∈velocitytimetaken

it has direction, the same as the change of velocity

a = 3 m/s2 means that velocity changes 3 m/s every second!!!If an object’s initial velocity is 4 m/s then after one second it will be 7 m/s, after two seconds 10 m/s,…

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► Acceleration can cause: 1. change in speed (speeding up: v nd a in the same direction;

slowing down: v and a in the opposite direction) 2. changing direction 3. both

ALL TOGETHER:

Any Motion

¿defition :vavg=xt→ x=vavgt

Motion with constant velocity – uniform motion

v = vavg at all times, therefore:

x = vt

In addition to these equations to solve a problem with constant acceleration you’ll need to introduce your own coordinate system, because displacement, velocity and acceleration are vectors (they have directions).

ALL WE KNOW ABOUT GRAPHS

Uniform Accelerated Motion equations x=vavgt always−for any motion

v=u+at

vavg=u+v

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x=ut+ a2t 2

v2=u2+2ax

Velocity at some time is equal to the slope of the tangent line at that position on the position-time graph.

Average velocity between two positions is equal to the slope of the secant line between these two points on the position-time graph.

Acceleration at some time is equal to the slope of the tangent line at that position on the velocity-time graph.

Average acceleration between two positions is equal to the slope of the secant line between these two points on the velocity-time graph.

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From velocity – time graph :

From acceleration – time graph :

► Free Fall formulas – Formulas are the ones for uniform accelerated motion with a = g

v = u + gt vavg = u+v

2 y = u+v

2 t y = ut + g2 t2 v2 = u2 + 2gy

g = 9.8 m/s2, downward ≈ 10 m/s2. Remember that in the coordinate system in which upward is chosen to be positive, g is negative and vice versa.If air resistance is not mention it is assumed that we ignore air resistance.When the object reaches maximum height, the velocity of the object is 0 m/s, but acceleration is still g = 9.8 m/s2

downward. Velocity changes, but g does NOT!!!

► Terminal speed/velocity – is maximum velocity an object can reach in air/any fluid.Air resistance depends on velocity. The greater velocity, greater air resistance, smaller acceleration.Acceleration is getting smaller due to air resistance and eventually becomes zero when the force of the air resistance equals gravity, the downward force of gravity is equal to the upward force of air resistance resulting in a zero net force, hence zero acceleration. The object will stop accelerating and maintain the same speed.Terminal velocity/speed is different for different bodies.

Area under the velocity-time graph between two times/positions is the displacement covered in that time interval.

Area under the acceleration-time graph between two times/positions is the change in velocityin that time interval.

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► Graphs of the motion of an object in the gravitational field (downward is positive)

PROBLEMS:1. A motorist travels 400 km at 80 km/h and 400 km at 100 km/h. What is the average speed of the motorist on this trip? A. 84 km/h B. 89 km/h C. 90 km/h D. 91 km/h E. 95 km/h

2. A car starts from rest and uniformly accelerates to a final speed of 20.0 m/s in a time of 15.0 s. How far does the car travel during this time?

A. 150 mB. 300 mC. 450 mD. 600 mE. 800 m

3. A car starts from rest and accelerates at 0.80 m/s2 for 10 s. It then continues at constant velocity. Twenty seconds after it began to move, the car has:

A. velocity 8.0 m/s and has traveled 40 m. B. velocity 8.0 m/s and has traveled 80 m. C. velocity 8.0 m/s and has traveled 120 m. D. velocity 16 m/s and has traveled 160 m. E. velocity 16 m/s and has traveled 320 m.

4. A racing car traveling with constant acceleration increases its speed from 10 m/s to 30 m/s over a distance of 80 m? How long does this take?

5. A rocket near the surface of the earth is accelerating vertically upward at 10 m/s2 . The rocket releases an instrument package. Immediately after release the acceleration of the instrument package is:

A. 20 m/s2 up B. 10 m/s2 up

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C. 0 D. 10 m/s2 down E. 20 m/s2 down

6. A racing car traveling with constant acceleration increases its speed from 10 m/s to 30 m/s over a distance of 80 m? How long does this take?

7. An object with an initial velocity of 12 m/s west experiences a constant acceleration of 4 m/s2 west for 3 seconds. During this time the object travels a distance of:

8. An object starts from rest and accelerates uniformly in a straight line in the positive x direction. After 11 seconds, its speed is 70.0 m/s. a. Determine the acceleration of the object. b.How far does the object travel during the first 11 seconds? c.What is the average velocity of the object during the first 11 seconds?

9. A brick is dropped from rest from a height of 4.9 m. How long does it take for the brick to reach the ground?

10. What maximum height will be reached by a stone thrown straight up with an initial speed of 35 m/s?

11. An object dropped from a stationary balloon hits the ground in 12.0 s. If its acceleration is 9.80 m/s2, the height of the balloon is:

12. How long does it take for a stone dropped off a 175-m high building to land on the ground ?

13. A ball is thrown from the edge of 80 m high cliff, upward at speed of 30 m/s. At what two times are the displacements 10 m upward? What is velocity at these times? At what two times are the displacements 10 m downward? What is velocity at these times?

14. When a falling object reaches terminal velocity, it: A. is no longer subject to the friction of air.B. moves downward with constant velocity. C. has an acceleration of approximately 10 m/s2. D. has no downward velocity. F. has an upward acceleration.

15. When a falling object moves with terminal velocity, it A. has zero velocity. B. has zero acceleration. C. has an upward acceleration. D. is no longer subject to air resistance. E. has an acceleration of approximately 10 m/s2 .

1. B 2. A 3. C 4. 4.0 s 5. D 6. 4.0 s 7. 54 m 8. a.+6.4 m/s2 b. 390 m c. +35 m/s9. 1.0 s 10. 62 m 11. 706 m 12. 6 s 14. B 15. B

GRAPHS OF 1 – D MOTION

PROBLEMS

1. Given the graph of the velocity vs. time of a chick flying due south for the winter. At what B point did the duck stop its forward motion?

A. A B. B C. C D. D E. none of these points

2. The graph represents tile relationship between distance and time for an object that is moving along a straight line. What is the instantaneous speed of the object at t = 5.0 seconds?

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A. 0.0 m/sB. 0.8 m/sC. 2.5 m/sD. 4.0 m/sE. 6.8 m/s

Between what times did the object have a non-zero acceleration?A. 0 s onlyB. 0 s to 5 sC. 5 s to 8 sB. 0 s to 8 sE. the object was not accelerating at any time.

3. The motion of a circus clown on a unicycle moving in a straight line is shown in the graph.

a. What would be the acceleration of the clown at 5 s?A. 1.6 m!s2 B. 2.0 m!s2 C. 3.4 m!s2 D) 8.O m!s2

E) none of the above

b. After 12 seconds, how far is the clown from her original starting point?

A. O m B. 10 mC. 34 m D. 47m E. 74m 4. The accompanying graph describes the motion of a marble on a table top for 10 seconds.

a. For which time interval(s) did the marble have a negative velocity?

A. from t = 8.0 s to t = 10.0 s onlyB. from t = 6.9 s to t = 10.0 s onlyC. from t = 4.8 s to t = 10.0 s onlyD. from t = 4.8 s to t = 6.2 s and from t = 6.9 s to t = 10.O s onlyE. from t = 3.2 s to t = 3.6 s, from t = 4.8 s to t = 5 s, and from t = 6.8 s to t= 7.2 s only.

b. For which time interval(s) did the marble have a positive acceleration?

A. from t = 8.0 s to t = 8.0 s onlyB. from t = 0.0 s to t = 3.6 s onlyC. from t = 3.8 s to t = 4.8 s and from t = 6.2 s to t = 6.8 s onlyD. from t = 2.0 s to t = 2.5 s, from t = 5.8 s to t = 6.2 s, and from t = 8.4 s to t = 8.8 s only.E. from t = 3.2 s to t = 3.7 s, from t = 4.8 s to t = 5.0 s, and from t = 6.8 s to t= 7.2 s only.

c. Vhat is the marbles average acceleration between t = 3.3 s and t = 3.8 sA. -2.0 m/s2 B. 0.0 m/s2 C. 0.8 m/s2 D. 2.0 m/s2 E. 3.0 m/s2

5. The accompanying graph describes the motion of a toy car across the floor for 10 seconds.

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a. What is the acceleration of the toy car at t = 4 s?

A. -1 m/s2 B. 0 m/s2

C. 1 m/s2

D. 2 m/s2

E. 4 m/s2

b. What was the total displacement of the toy car for the entire 10 second interval shown?A. 0 meters B. 6.5 meters C. 9 meters D. 10 meters E. 11.5 meters6. Graph refers to the motion of a toy car traveling along the x-axis. It is a plot of the car’s velocity in the x direction.

a. During what time interval was the car moving towards its initial position at constant velocity? A. 0-1Os B. 10-20s C. 20-25 s D. 25-30 s E. 30-35 s b. What was the acceleration at 33 s? A. +0.40 m/s2 B. +0.20 m/s2

C. 0D. -0.20 m/s2

E. -0.40 m/s2

c. How far did the car travel during first 15 seconds?A. 0B. 3.0 mC. 15 mD. 30 mE. 45 m

1. C 2. B, C 3. B, E 4. D, D, A 5. C, B 6. D, B, D

► Uniform Circular Motion – An object is moving in circular path at constant speed.Average velocity is zero, because displacement for a full circle is zero.velocity – changing direction - tangent to the path (circle) and constant magnitude (speed) centripetal acceleration – points toward the center and constant magnitude⇒ velocity and centripetal acceleration vectors are always perpendicular to each other

speed v = distance/time v = 2πrT r – radius of the circle period T is time required for one complete revolution

Centripetal acceleration that maintains motion around the circle with radius r, with speed v is given

ac=v2

r (ac)=m /s2

PROBLEMS1) A girl sits on a tire that is attached to an overhanging tree limb by a rope. The girl's father pushes her so that her centripetal acceleration is 3.0 m/s2. If the length of the rope is 2.1 m, what is the girl's speed?

2) A rock tied to a string is traveling at a constant speed of 4 m/s in a circle of radius 1.5 m. Calculate the magnitude of the centripetal acceleration of the rock. What is the direction of the acceleration?

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3. A rock tied to the end of a string moves in a circle at a constant speed of 2.5 m/s and experiences an acceleration of 4.0 m/s2. What is the radius of the circle of its motion?1) v=2.5m/s 2) ac = 10.7 m/s2 3) r = 1.56 m

Projectile motion

Problem:1. A kid is playing with the pebbles at the top of a beautiful cliff above the sea. He flicks a pebble horizontally with the speed of 10 m/s. The pebble hits the sea 3s later. a. How high is the cliff above the sea? b. Find the speed and velocity 2s after it was thrown. Find the speed and velocity just before it hits the sea. c. Draw graphs: displacement vs. time, velocity vs. time, acceleration vs. time for each component of the motion. Write equation in each case.

2. In order to appreciate variation of speed and velocity during projectile motion, we calculate the values of a projectile velocity and speed for successive seconds t = 1s, 3s, and 5s, which is projected with an initial velocity of 60 m/s making an angle of 300 with the horizontal. Is that object on the way up or on the way down at those times. Find position (x,y) at the same times. How long does it take for that object to reach the same height? What is the maximum height? Draw graphs: displacement vs. time, velocity vs. time, acceleration vs. time for each component of the motion. Write equation in each case.

HORIZONTAL MOTION VERTICAL MOTIONux = u cos 0 uy = u sin 0

vx = ux vy = uy + gt

x = ux t