TEST 2 Q 1 some ACADEMIC review questions to try
Define: displacement, velocity, average velocity, average speed, acceleration.
Displacement: change in distance from start (with direction)
Velocity: change in displacement over change in time, (indicates direction)
Average velocity: total change in displacement over change in time, the average displacement for each time interval
Average speed: the average distance traveled for each time interval, total distance traveled over change in time.
Acceleration: the increase in velocity over each time interval, change in velocity over change in time.
Describe the motion of an object that starts with:
Positive velocity, positive acceleration it speeds up going forwards, displacement always increasing. à. . . . . . . . . . . . à
Positive velocity, negative acceleration it starts going forwards, getting slower, but still going forwards, eventually stopping for a split second then reversing direction and going backwards. à. . . . . . . . . . . . à
ß. . . . . . . . . . . . ß↩
Negative velocity, positive acceleration it starts off going backwards fast, then gets slower while still going backwards, eventually stopping for a split second then reversing direction and going forwards.ß. . . . . . . . . . . . ß
➘à. . . . . . . . . . . . à
Negative velocity, negative acceleration It speeds up going backwards, displacement from start is always decreasing (getting farther away in the reverse direction) ß. . . . . . . . . . . . ß
When does an object have an average velocity of zero?When the displacement is zero (it returns to its starting point)… note that its speed could be anything!
A ball is rolled down a ramp.
Sketch the D-T and V-T graph.
If down is positive
Explain how to get the acceleration, using only a stopwatch and ruler. (T & D)
If you know the initial velocity is zero, then the average velocity is D/T. the final velocity is twice the average (middle) velocity. Acceleration is change in velocity over time.
Or using equation #4: A = 2D/T2 if Vi=0
If the a is 1 m/s2, what is the ball’s velocity after 5 seconds? Vi=0, A=1, T=5, Vf=???
Vf=Vi+AT = 0+AT = AT = 1(5) = 5 m/s… the ball increases velocity by 1 m/s every second for 5 seconds…
What is its average velocity? If the ball has a final velocity of 5 m/s, and an initial velocity of 0 m/s, its average velocity is Vavg= (Vi+Vf)/2=(0+5)/2=5/2= 2.5 m/s
How much distance has it covered in that time? Dis is Vavg*T = 2.5 (5)=12.5 meters
Or D=ViT+ ½*AT2 = 0+ ½(1)(52)=1/2*25= 12.5 m
What was the distance traveled in the 5th second? (from 4 to 5)
The distance traveled in the 5th second is the distance at 5 sec minus distance at 4 sec….
Distance at 4 sec = D=ViT+ ½*AT2 = 0+ ½(1)(42)=1/2*16= 8 m, so distance in the 5th second is D(5)-D(4)= 12.5-8= 4.5 m
Or…. The average velocity from T=4 to T=5 is (Vf(4)+Vf(5))/2= 4.5 m/s in one second… D=Vavg*T=4.5*1=4.5 m
A ball is thrown up in the air and caught at the point of release.
Sketch the D-T and V-T graph.
Explain what two pieces of information would help calculate the maximum height.You know that at the maximum height the object has a Vf of 0, and the time is half the total time of the trip.
Explain what would be different about the ball’s path, its maximum height, its graphs, and its time and velocity if this were done on Jupiter (greater gravity).
The gravity is more on Jupiter, thus the ball would not go as high, and hit the ground sooner, the acceleration due to gravity would be different, but the final velocity at which it hits the ground would still be the same as the start….
A speeder going at a constant speed passes a police car, who starts accelerating from 0 some time later. Explain how to calculate when and where the cops catch the speeder, and sketch a strobe motion dot image, D-T, and V-T graph for each.
If the speeder is going at a constant speed, then his distance is rate times time… or D=VT . . . . . . . . . . . . . . . . . . . . .
If the police car is starting at Vi=0, then his distance is average velocity (1/2 final velocity=1/2AT) times time or D= 1/2AT2
. . . . . . . . . . . . .
They meet at the same place (D) at the same time (T) D=D, T=T, so set the two equations equal and solve for D and T…… VT=1/2AT2
TWO WORD PROBLEMS TO TRY (ACADEMIC):
I start traveling at 3 m/s and accelerate at the rate of 2 m/s2 up to a max of 25 m/s.
A tiger is 70 meters behind me traveling a constant 20 m/s.
When I reach my maximum speed of 25 m/s, will I be in front of, or behind the tiger?
For D, eq #5 Vf2=Vi2+2AD 252=32+2*2*D D= (625-9)/4=154 meters =Dme
For T, eq #3 Vf=Vi+AT 25=3+2(T) T=(25-3)/2=11 seconds
In that time…… Tiger Di =-70, Vi=Vf=Vavg=20m/s, A=0
I am barely in front of the tiger!!!
I stand on a cliff and jump up at 39.2 m/s. If I hit the valley below after 10 seconds, how fast am I going? How high is the cliff? How high did I jump?
Vi=39.2 m/s, A=-9.8 m/s2, T=10sec
Vf= Vi +AT = 39.2 + -9.8 * (10) = -58.8 m/s downward
D=ViT+1/2AT2 = 39.2 (10) + ½*-9.8*100 = -98 meters below start is cliff
How high? At the apex Vf = 0 m/s so:
Vi=39.2 m/s, Vf=0m/s, A=-9.8 m/s2, Dheight=????
eq #5 Vf2=Vi2+2AD 02=39.22+2*-9.8*D D= (-1536.64)/-19.6=78.4 meters =Dheight