Physics 221, January 21

Key Concepts:

•Scalar Quantities and Vector Quantities•Position, Distance, Displacement•Average Speed, Average Velocity•Instantaneous Speed, Instantaneous Velocity•Average Acceleration, Instantaneous Acceleration

Please find a seat.Keep all walkways free for safety reasons and to comply

with the fire code.

Please do not use social media during class.

Physics: the key word is change!Matter interacts and the interactions change the physical state of matter.The laws of physics predict when and how the physical state of matter changes.

The laws of mechanics predict when and how things move.We have to agree on a way to describe motion.We use vector quantities with magnitude and direction and scalar quantities with onlymagnitude to describe motion.

Vector quantities:Position: r = ai + bj + ckDisplacement: ∆r = (x2 - x1)i + (y2 - y1)j + (z2 -z 1)kAverage velocity: v = ∆r/∆t. Instantaneous velocity: Let ∆t 0.Average acceleration: a = ∆v/∆t.Instantaneous acceleration: Let ∆t 0.

Scalar quantities: distance, average speed, instantaneous speed

*Which of the following statements about vectors and scalars are TRUE?

1. A vector is a large quantity and a scalar is a small quantity.

2. A vector quantity has a direction and a scalar does not.

3. The quantity 20 m/s, north is a speed and as such is a scalar quantity.

4. The quantity 9.8 m/s2 is an acceleration value and as such is a vector quantity.

5. A scalar quantity has a magnitude and a vector quantity does not.

Distance and displacementA person walks from A to B along a circular path of radius 10.00 m around 1/2 of the circle. What is the magnitude (in m) of the displacement vector?

A. 0 mB. 62.8 mC. 10 mD. 5 mE. 31.4 mF. 20 m

Speed and velocityFour different mice (labeled A, B, C, and D) ran the triangular maze shown below. They started in the lower left corner and followed the paths of the arrows. The times they took are shown below each figure, (i) Which mouse had the greatest average speed?(ii) Which mouse had the greatest average velocity magnitude?

1. (i) D, (ii) C 2. (i) B, (ii) A 3. (i) A, (ii) D 4. (i) B, (ii) C

Hint:speed = total distance traveled/time|velocity|= |displacement|/time

Example: Mouse Cspeed = 2L/(4 s)|velocity|= L/(4 s)

L = length of one side of the triangle

Motion in 1 dimensionIf an object is restricted to move in one dimension, for example along the x-axis, we can specify the vector quantities

position, displacement, velocity, and accelerationby a signed number with units. The sign of the number than specifies the direction.

In one dimension, if the x-component of a vector is positive, the vector ispointing in the positive x-direction, and if the x-component of a vector is negative, the vector is pointing in the negative x-direction.

Speed and velocityA man walks his dog on a very straight path in a park. He can walk back and forth as slow or fast as he wants to, but he has to stay on the path. Which of the following statements is correct for an arbitrary time interval?

1. His average speed must be equal to the magnitude of his average velocity.

2. His average speed can be less than the magnitude of his average velocity.

3. His average speed can be greater than the magnitude of his average velocity.

4. Since he is walking along a straight path, his average acceleration must be zero.

5. His average velocity must be zero.

Hint:Is the man required to always travel in the same direction on the straight line path?

Relative velocity

The velocity of an object depends on the reference frame, it is not a quantity all observers agree on.

An object can be at rest in one reference frame, and having non-zero velocity in another reference frame.

When a parent tells a child to sit still in a moving car, they mean to sit still with respect to the car, not with respect to the road.

Relative velocityThe water of a river flows with a velocity of +2 m/s in the positive x-direction with respect to the shore. A boat as a velocity of +7 m/s in the positive x-direction with respect to the shore. What is the velocity of the boat with respect to the water?

1. +9 m/s2. +7 m/s3. +5 m/s 4. -2 m/s5. -5 m/s

AccelerationUnderstanding acceleration is the key to doing well in this course.

a = ∆v/∆t = (vf – vi)/(tf – ti)The acceleration a is a vector. It is the rate at which the velocity vector is changing. The velocity vector changes when its magnitude changes or its direction changes.

Whenever your velocity is CHANGING, you are accelerating.

You are accelerating when you CHANGE your speed, CHANGE your direction of travel, and when you change both.The keyword is CHANGE.

constant velocity < -- > zero acceleration

*Which one of the following situations is not possible?

1. A body has zero velocity and non-zero acceleration.

2. A body travels with a northward velocity and a northward acceleration.

3. A body travels with a northward velocity and a southward acceleration.

4. A body travels with a constant acceleration and a time-varying velocity.

5. A body travels with a constant velocity and a time-varying acceleration.

A 50-g superball traveling at 20.0 m/s bounces off a brick wall and rebounds at 18.0 m/s. A high speed camera records this event. If the ball is in contact with the wall for 2*10-3 s, what is the magnitude of the average acceleration during this time interval?

1. 9.5 m/s2

2. 25000 m/s2

3. 19000 m/s2

4. 1000 m/s2

5. 0.001 m/s2

Hint:a = (vf – vi)/∆t∆t is very small (milliseconds).

vf and vi are vectors and have opposite signs.

The magnitude of a is a positive number.

The direction of the acceleration vector

Consider motion in one dimension.

When do the velocity and the acceleration vector have the same direction?

When do the velocity and the acceleration vector have opposite direction?

Can an object that is momentarily at rest be accelerating?

A “motion diagram” is shown. It is a series of snapshots of a bunny’s position taken every 1.0 seconds. What is the direction of the acceleration vector if the bunny ismoving to the left?

1. In the negative x-direction2. In the positive x-direction3. The acceleration is zero.4. The acceleration does not have a

direction.

Acceleration is absolute

The acceleration of an object does not depend on the reference frame, it is a quantity all observers agree on.

Even without outside references you can determine if you are accelerating.

Example: an elevator ride.You momentarily feel heavier when the elevator accelerates upward and lighter when it accelerates downward.

Motion Graphs for 1D motionWe can represent one-dimensional motion using a position versus timegraph or a velocity versus time graph.

Position versus time graph:The slope gives the instantaneous velocity. (vx = ∆x/∆t)

Positive slope positive velocityNegative slope negative velocityConstant slope constant velocityChanging slope acceleration(The position versus time graph of motion with constant acceleration is a section of a parabola.)

Velocity versus time graph:The slope gives the instantaneous acceleration. (ax = ∆vx/∆t) Positive slope positive accelerationNegative slope negative accelerationConstant slope constant acceleration

Getting information from graphs:

The position versus time graph represents the motion of an object moving along a straight line. What is the instantaneous velocity of the object at t = 14 s?

1. -(1/15) m/s 2. -(5/3) m/s3. 0.067 m/s4. (1/3) m/s5. -(2/3) m/s

The velocity versus time graph represents the motion of an object moving in a straight line. Is the magnitude of the average velocity equal to the average speed between t = 0 and t = 15 s?

1. Yes2. No

The velocity versus time graph to the right represents the motion of an object moving in one dimension. What is the object's average acceleration between t = 0 andt = 6 s?

1. -3 m/s2

2. 1.5 m/s2

3. -0.83 m/s2

4. 0.67 m/s2

5. 0.5 m/s2

The positions of two blocks at successive 0.20-second time intervals are represented by the numbered squares in the figure below. The blocks are moving towards the right. The accelerations of the blocks are related as follows:

1. The acceleration of a is greater than the acceleration of b.

2. The acceleration of a equals the acceleration of b. Both accelerations are greater than zero.

3. The acceleration of b is greater than the acceleration of a.

4. The acceleration of a equals the acceleration of b. Both accelerations are zero.

5. Not enough information is given to answer the question.

Velocity versus time graphs are shown for 4 objects, A, B, C, and D. Which statement is true?

1. Sometime between t = 16 and 17 s the objects have the same acceleration.

2. Object A is at rest.3. All objects are moving with constant

velocity.4. None of the above.

Hint: ax = ∆vx/∆t (slope)

Extra problems

The problems below are solved using the kinematic equations for one-dimensional motion with constant acceleration.

1D kinematic equations (constant acceleration)

Let ∆t = tf – ti.Thenvxf = vxi + ax∆t or ∆vx = ax∆t,

vx(avg) = (vxf + vxi)/2,

xf = xi + vxi∆t + ½ax∆t2 or ∆x = vxi∆t + ½ax∆t2.

Combining these equations we getvxf

2 = vxi2 + 2ax(xf - xi).

A racecar has a speed of +90 m/s when the driver releases a drag parachute. If the parachute causes a uniform deceleration of –15 m/s2, how long will it take the racecar to stop?

1. 15 s2. 6 s 3. 4 s4. 10 s5. 9 s

Solution:Since the velocity decreases uniformly, theacceleration is constant. We therefore havea = - 15 m/s2= (vf – vi) /(tf – ti)

= (0 m/s – 90 m/s)/(∆t ).

∆t = (90 m/s)/(15 m/s2) = 6 s.

A racecar has a speed of +90 m/s when the driver releases a drag parachute. If the parachute causes a uniform deceleration of –15 m/s2, how far will the car travel before it stops?

1. 270 m 2. 30 m3. 9 m4. 600 m5. 810 m

Solution:Since the velocity decreases uniformly, theacceleration is constant. We therefore havea = - 15 m/s2.

vf2 = vi

2 + 2a(xf – xi).0 = (90 m/s)2 – (30 m/s2) ∆x.∆x = (8100/30) m = 270 m.

A particle is moving with velocity v = 60 m/s at t = 0. Between t = 0 and t = 15 s the velocity decreases uniformly to zero. What was the acceleration during this 15 s time interval?

1. 4 m/s2

2. 8.6 m/s2

3. 0.53 m/s2

4. -0.27 m/s2

5. -4 m/s2

Solution:Since the velocity decreases uniformly, the acceleration is constant. We therefore have

a = (vf – vi) /(tf – ti) = (0 m/s – 60 m/s)/(15 s)= - 4m/s2

Let the x-axis of your coordinate system point along a drag strip. A drag racer starts from rest and accelerates uniformly to a speed of 76.5 m/s in 8.5 seconds. What is the acceleration of the car?

1. 2.1 m/s2

2. 68 m/s2

3. 9 m/s2

4. 650 m/s2

5. 76.5 m/s2

Solution:Since the velocity increase uniformly, the acceleration is constant. We therefore have

a = (vf – vi) /(tf – ti) = (76.5 m/s – 0 m/s)/(8.5 s)= 9 m/s2

Let the x-axis of your coordinate system point along a drag strip. A drag racer starts from rest and accelerates uniformly to a speed of 76.5 m/s in 8.5 seconds. What is the distance traveled by the carin the 8.5 s?

1. 688 m 2. 38 m 3. 76.5 m 4. 650 m5. 325 m

Solution:Since the velocity increase uniformly, the acceleration is constant. In the previous problem we found that a = 9 m/s2.

Now use∆x = vxi∆t + ½ax∆t2

= 0*8.5 s + ½(9 m/s2)*(8.5 s)2

= 325 m.