Wednesday, September 19, 2012

Physics Classroom Kinematics Practice


Physics Classroom Kinematics Practice
Michael Matias
September 19th, 2012

1.       An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
d= v_i*t+1/2*a*t^2→d=0+1.6*(32.8)^2→d= 1721m
2.       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+1/2*a*(5.21)^2→110=13.57a→a=8.1m/s^2 
3.       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?
v_f=v_i+a*t→v_f=0-9.8*2.6→v_f=-25.48m/s
d=(v_i+v_f)/2*t→d=(-25.48)/2*2.6→d=-33.1m 

4.       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→27.6=2.47a→11.17m/s^2 
d=(v_i+v_f)/2*t→d=(18.5+46.1)/2*2.47→d=79.79m 
5.       A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.
d= v_i*t+1/2*a*t^2→1.4=0+.8*t^2→t^2=1.75→t=1.32s 
6.       Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance that the sled travels?
v_f=v_i+a*t→444=1.8*a→a=246.67m/s^2 
d=(v_i+v_f)/2*t→d=444/2*1.8→d=399.6m  
7.       A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.
d=(v_i+v_f)/2*t→35.4=3.55t→t=9.97s 
v_f=v_i+a*t→7.71=0+9.97*a→a=.77m/s^2 
8.       An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?
v_f^2=v_i^2+2ad→〖65〗^2=0+2*3*d→4225=6d→d=704.2m 
9.       A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration).
d=(v_i+v_f)/2*t→d=22.4/2*2.55→d=28.56m 
10.   A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.
v_f^2=v_i^2+2ad→0=v_i^2+2*-9.8*2.62→v_i^2=51.352→v_i=7.2m/s 
11.   If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?
v_f^2=v_i^2+2ad→0=v_i^2+2*-9.8*1.29→v_i^2=25.284→v_i=5.03m/s  
Time to peak -> d=(v_i+v_f)/2*t→1.29=5.03/2*t→t=.51s
Hang Time = 2*.51 = 1.02s


12.   A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).
v_f^2=v_i^2+2ad→〖521〗^2=0+2*a*0.84→a=161572m/s^2 
13.   A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the total hang-time.)
v_f=v_i+a*t→0=v_i-9.8*3.125→v_i=30.625m/s  
d=(v_i+v_f)/2*t→d=30.625/2*3.125→d=47.9m 
14.   The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.
d= v_i*t+1/2*a*t^2→370=0+1/2*9.8*t^2→370=4.8t^2→t=8.78s 
15.   A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)
v_f^2=v_i^2+2ad→0=〖367〗^2+2*a*.0621→0=134689+0.1242a→a=-1084452m/s^2 
16.   A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.
d= v_i*t+1/2*a*t^2→d=0+1/2*-9.8*〖3.41〗^2→d=-57m 
17.   It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.
v_f^2=v_i^2+2ad→0=v_i^2+2*-3.9*290→v_i^2=2262→v_i^2=47.6m/s 
18.   A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.
v_f^2=v_i^2+2ad→〖88.3〗^2=0+2*a*1365→a=2.86m/s^2 
d=(v_i+v_f)/2*t→1365=88.3/2*t→t=30.92s 
19.   A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (assume uniform) of the dragster.
v_f^2=v_i^2+2ad→〖112〗^2=0+2*a*398→a=15.76m/s^2 
20.   With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.
v_f^2=v_i^2+2ad→0=v_i^2+2*-9.8*91.5→v_i^2=42.4m/s=94.6mi/hr 

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

Monday, September 10, 2012

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

Monday, September 3, 2012

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


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


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