Test review for finding average speed and average velocity

For each of the following cases, find average speed and average velocity
1. An object moves 20mto the left, then 50m to the right then again 100m left. The trip took 20s.

answers: Average speed: 8.5m/s Average velocity:3.5m/s to the left

Average Speed = 170/20 = 8.5m/s
Average Velocity = 70/20 = 3.5m/s left

2. An object moves west 40m, back east 5m , then north 15m. Trip took 2.5 minutes.
Answers: Average speed: 0.4m/s Average velocity: 0.25m/s at 23.2⁰ off the west. )

Average Speed = 60/2.5 = 24meters/minute = 0.4m/s
Average Velocity = 38.1/2.5 = 0.25m/s NW 23.2 degrees

Physics Classroom Practice

A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

d= v_i*t+1/2*a*t^2→110=0+.5a*27.14→.5a=4.05→a=8.1m/s^2

Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall?

d= v_i*t+1/2*a*t^2→d=0+.5*-9.8*6.76→d=32.68m

A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

v_f=v_i+a*t→46.1=18.5+a*2.47→a=11.17m/s^2

d=(v_i+v_f)/2*t→d=79.78m

Kinematics Review

Kinematics equations/ review guideline

Chapter 3

1.         A tennis ball is dropped from 1.2m above the ground. It rebounds to a height of 1m.

a.        With what speed does it hit the ground?

d= v_i*t+1/2*a*t^2→1.2=0*t+1/2*9.8*t^2→4.9t^2=1.2→t^2=2.4→t=1.55s

v_f=v_i+a*t→ v_f=0+9.8*1.55=15.19m/s 

b.        With what speed does it leave the ground?

v_f^2=v_i^2+2*a*d→0= v_i^2+2*-9.8*1→v_i^2=19.6=4.4m/s

2.         A car starts from rest and gets to a store that is 400m away. The car’s velocity when it gets to the store is 10m/s.  Another car that starts in the same place, is already moving at 12m/s. Which car is going to get to the store first and by how many seconds less will it travel?

CAR 1: d=(v_i+v_f)/2*t→400=(0+10)/2*t→400=5t→t=80s

CAR 2: d=(v_i+v_f)/2*t→400=12t→33.3s

CAR 2 will get to the store 46.66s before the CAR 1

3.         Engineers are developing new types of guns that might someday be used to launch satellites as if they were bullets. One such gun can give a small object a velocity of 1m/s moving it through only 2cm.

a.        What acceleration does the gun give this object?

d=(v_i+v_f)/2*t→.02=(0+1)/2*t→t=.04s 

v_f=v_i+a*t→1=0+a*.04→1=.04a→a=25m/s 

b.        Over what time interval does the acceleration take place?

d=(v_i+v_f)/2*t→.02=(0+1)/2*t→t=.04s 

4.         A rock is thrown down into a well with an initial downward speed of 3m/s. The rock hits the bottom 1.8s later.

a.        How deep is the well?

v_f=v_i+a*t→v_f=3+9.8*1.8→v_f=20.64m/s 

d=(v_i+v_f)/2*t→d=(3+20.64)/2*1.8→d=21.276m 

b.        What is the rock's speed as it hits the bottom?

v_f=v_i+a*t→v_f=3+9.8*1.8→v_f=20.64m/s 

5.         A hot air balloon is ascending at a steady velocity of 2.2m/s. An object is dropped from it.  What is the distance from the object and the balloon after 0.5s?

d= v_i*t+1/2*a*t^2→d=2.2*.5+1/2*9.8*〖.5〗^2→d=2.325m

Motion Exercises

Motion exercises

1.    A car moving at 36km/h is uniformly accelerated at the rate of 0.5m/s.

a.    What is the speed of the car 6s after the acceleration begins?

v_f=v_i+a*t→ v_f=10.83+.5*6=13.83m/s 

b.    How far does the car move during the 6s

d=(v_i+v_f)/2*t→d= (36+39)/2*6=225m 

2.    A ball rolls down a hill accelerating at 3m/s.

a.    How far did it move in 4s if it starts from rest?
     v_f=v_i+a*t→v_f= 0+3*4=12m/s
     d=(v_i+v_f)/2*t→d=(0+12)/2*4=24m

b.    What is its speed after the 4s?

v_f=v_i+a*t→v_f= 0+3*4=12m/s

3.    Pressing the brake of a car it slows down from 30m/s and comes to stop during 6s.

a.    What is its stopping distance?

d=(v_i+v_f)/2*t→d=(30+0)/2*6=90m

b.    What is its acceleration?

 v_f=v_i+a*t→0=30+a*6→ -30=6a→a=-5m/s^2

4.    Two cars are moving at 30m/s. One car starts accelerating at 0.2m/s.(the other car continues in the same speed). What will the distance between the two cars be 4s after the acceleration started?

Car 1: v_f=v_i+a*t→v_f=30+.2*4=30.8m/s^2

            d=(v_i+v_f)/2*t→d=(30+30.8)/2*4=121.6m

Car 2: d=(v_i+v_f)/2*t→d=(30+0)/2*4=60m

Distance between the two: 121.6-60 = 51.6m

5.    An object moving at 20m/s starts accelerating at -0.5m/s covering a distance of 20 m. How long does this motion take?

d= v_i*t+1/2*a*t^2→20=20*t+1/2*-0.5t^2→0=-0.25t^2+20t-20→t=1.01s

Constant Velocity

Kinematics activity

Constant velocity

Use the 3 colored tubes to analyze the motion of an air bubble.

For each tube – start with the bubble on the bottom. Then Mark on the tube the position of the bottom of the bubble after each second as it is rising up.

Measure the displacement of the bubble at the end of 1,2,3,4….. (as many seconds as you have).

On the same axes, draw a graph of position vs time for each tube. Label on the line which color it is. Use proper graphing techniques.

Answer the following questions:

1.     What is the slope in each case (including units) and what does the slope indicate?

2.     In which tube was the bubble fastest? Slowest? How can you tell?

3.     Show two examples of extrapolation with the green tube.

4.     Show 2 examples of interpolation with the red tube.

5.     How can we tell based on the graph that the bubbles in all tubes were moving at a constant velocity?

Average Speed and Average Velocity

Finding average speed and average velocity

Remember: Average speed= distance/time

                    Average velocity = displacement/time

 For each one of the following cases, find the average speed and average velocity, both in m/s

1.       Moving 50m east then 70mwest in 2 minutes.

 Average Speed = 120meters/2minutes = 60meters/minute = 1m/s

Average Velocity = 20meters/2minutes = 10meters/minute = .17m/s west

2.       Moving north 60m, south 80m and north again 10m in 4.5minutes.

Average Speed = 150meters/4.5minutes = 33.33meters/minute = 0.56m/s

Average Velocity = 10meters/4.5minutes = 2.22meters/minute = .04m/s south

3.       Moving 60 m east in 2 hours

Average Speed = 60meters/2hours = 30meters/hour = .008m/s

Average Velocity = 60meters/2hours = 30meters/hour = .008m/s east

4.       Moving north 50m then east 20m in 1.5 minutes

Average Speed = 70meters/1.5minutes = 46.7meters/minute = .78m/s

Average Velocity = 53.85meters/1.5minutes = 35.9meters/minute = .6/s 21.8 degrees northeast 

5.       Moving east 30m, west 20m then south 70m in 2 minutes.

Average Speed = 120meters/2minutes = 60meters/minute = 1m/s

Average Velocity = 70.71meters/2minutes = 35.36meters/minute = .59m/s 81.9 degrees southeast 

Unit 1 Review

Mathematical introduction to physics

1.     Express  a:     v=  square root of  2as

v² = 2as à v²/2s = a

2.     Express t:      v =  v₀  +  at

(v-v₀)/a = t

3.     Express  c:      E  =  mc²

E/m = c² à sqrt(E/m) = c

Solve the following (include units)         N = Kgm/s²    j = Nm   W=j/s

                                   

4.     2.5x10¹² m/s²  x   3.3x10¹³ s  = (8.25x10²⁵)m/s

5.     3.5x10^-4kg  x  2.5x10^-10m/s² = (8.75x10^-14)N

6.     6.5x10⁵m/s  :    3.3x10^-3s  = (1.97x10⁸)m/s²

7.     2x10⁵N    :     3x10¹⁵m/s² = (6.67x10^-11)Kg

8.     2.4x10⁶w  *  2.3x10³s  = (5.52x10⁹)j

Can these equations be correct?  Use dimensional analysis of units to prove your point!   d (m)     v (m/s)      t (s)     a (m/s2)

1.     d = v/t   à m = (m/s)/s à m = m/s² NOT VALID    

2.     d = vt + at²/2 à m = (m/s)(s) + (m/s²)(s²)/2 à m = m + m/2 VALID

3.     v = 2ad à (m/s) = 2(m/s²)(m) à (m/s) = (m²/s²) NOT VALID

                                                                                     

Solve (your answer should be stated in MKS units)

1.     5.3x10³km + 20m  = 5300000m + 20m = 5300020m                   

2.     0.5m/s + 2km/h  =  0.5m/s + 0.56m/s = 1.06m/s

3.     5.5x10³ g + 20 kg = 5.5kg + 20kg = 25.5kg                          

4.     42cm + 2m  = 0.42m + 2m = 2.42m

 

Find the angles         a                               a= 15         b= 22

                                               b

∠A = tan^-1(15/22) = 34.3⁰                                              

c = 26.6

∠B = cos^-1(15/26.6) = 55.7⁰               b                                

                                                                         25⁰       

Find the sides a and  b          a                    

           25

a à sin(25⁰) = a/25 à a = 10.6

b à cos(25⁰) = b/25 à b = 22.7

Use units and dimensional analysis to find out if the formulas are valid

1.     Distance = speed * time à m = (m/s)*s à VALID                         

2.     acceleration = speed/time à (m/s²) = (m/s)/(s) à VALID

3.     Force = mass * acceleration à N = kg*(m/s²) à VALID                    

4.     distance = energy * time à m = (j)*(s) à NOT VALID

5.     Power = work * time à (j/s) = (j)*(s) à NOT VALID                              

6.     speed = power/force à (m/s) = (j/s)/(N) à VALID

The displacement of an object from a start point was checked at 2 s intervals. The results are in the following chart

Time (t) in (s)	Displacement (d) in (m)
0	0
2	4
4	8.5
6	12.2
8	15.8
10	21
12	25.3

a.     Graph the results using proper graphing techniques.

b.     What is the relationship you observe?

c.      Find the constant (slope) including units. What quantity does it represent?

d.     Write the equation of the line.

e.      Show one example of interpolation and one for extrapolation.