Topic: hw2a #1
A honeybee leaves the hive, flies in a straight line to a flower 5 km away in 10 min, and then takes 10 minutes to return (also in a straight line).
a.) Please find the distance travelled and displacement for the entire trip:
b.) Please find the average speed and average velocity for the entire trip:
Student A: For some reason I am not understanding the displacement principle. I understand
that you subtract the final position from the inital position. Its position goes from
zero to five and then five to zero, I seem to keep missing this little error! Anything
im not looking at? Thanks!
Dr. Man: Recall that i asked a similar question during class. On the blackboard I drew the
trace of someone running along a circle for many many times and ended at where he started.
What was the displacement during that WHOLE motion? Back to concepts and equations,
for this questions, the whole motion they are asking about, where is the real initial position
and where is the real FINAL position? To read more test book before and after lectures
can help.
Topic: HW 2a, #1, part c
Student A: Can anyone give me any hints? THANKS!
Dr. Man: Try to understand the definition of average speed or average velocity.
averge speed = what / time?
averge velocity = what / time?
For a unknown motion not flown in a straight line, (you only know it started here, went
to the flower and backed here), Can you decide its total distance of travel? can you decide
it's displacement of the whole motion?
Topic: HW#2a # 2
The average velocity of an object for a given motion points to the right.
When is it possible for the instantaneous velocity of the object to point to
the left? (Select all that apply.)
Student A: I'm confused with this question. I thought it was at the beginning of the motion, but
I was wrong. Then I thought it it was at the middle and the beginning and that's
wrong too. Can someone explain this?
Student B: I don't know if you have had your lab yet but if you have think of the motion
sensor and what the graph looked like when you walked in a negative direction. the
graph doesn't phyically point left it phyically points down, draw a picture and see if
that helps. Look at the overall slope.
Dr. Man: What student B says is that in the position vs t plot, when you go along the
negative direction, (toward the sensor), position vs time plot has a negative slope
("points down"). Student B, actually for that gragh, we don't expect anything to "point left".
You know the horizontal axis is time. In that graph, toward right time increases, toward
left time decreases. We are always looking at as time increases, whether position is moving
away from origin (along the positive position direction) or closer to the origion (along the
negative direction). Moving away from origin, gives you positive velocity (positive slope)
Moving toward origin gives you negative slope (negative velocity)
Back to the question Student A asks:
They key is to understand the definition of average velocity:
average velocity= total DISPLACEMENT/ time.
average velocity points to the right meaning the TOTAL displacement over the whole period
time is toward right. Now they ask is it possible for the instantaneous velocity of the object to
point to the left. And you should include all the possible choices which works.
Topic: HW #2a #4
Question: At the starting gun, a runner accelerates at 1.9 m/s2 for 4.0 s. The runner's
acceleration is zero for the rest of the race.
(a) What is the speed of the runner at t = 2.1 s?
Student A: I cant figure out what equation to use to find the speed. Can anyone help?
Dr. Man: Please list the question number as well. So that other people can help you better.
Read my lecture notes.
a, v_0, x_0, t, v_f
which ones are known? What is asked in question (a)? Which one is not known and also
not asked so that not needed? Choose the equation that doesn't have the one not being asked.
Topic: HW 2a #3
You drive at 20.0 m/s for 50 km and you drive for 40 m/s for 50 km
Is your average speed > = or < than 30 m/s?
Dr. Man: This is a very tricky quetsion.If you read chapter 2, you know the answer.
In the class we talk about CONSTANT ACCELERATION motion.
suppose that your velocity increased from 20m/s to 40m/s at a CONSTANT ACCELERATION,
your average velocity is for sure the average of 20 and 40
V_ave= (V_0+V_f)/2 = 30 m/s why?
Because that your velocity change is at constant rate. for the same amount of time, you
change the same amount of velocity.
When you look at V vs t plot of velocity increased from 20m/s to 40m/s at a CONSTANT
ACCELERATION, you see a straight line raising up w/ positive slope from 20m/s to 40m/s.
The middle average is exactly 30 m/s, because you spend the same amount of TIME travel
at velocity lower than 30m/s as you spend travel at velocity higher than 40m/s.
You can always understand this better if it is money.
If your salary increase at a constant rate from 20$ per day to 40$ per day. (suppose your
salary increase 1$ per day). At the 21st day It is as if your average salary is 30$ per day.
Right? the total money you made would be 620. You spent the same amount of time
earning less than 30 and earning more than 30....But question 3 in HW 2a is different.
It is NOT CONSTANT ACCELERATION motion.It only had two speed. This question
is about TWO parts of CONSTANT SPEED motion. But think about this. this person
spend more time drving with the low speed or the high speed. (read book for more
details about the same question).
Topic: HW 2a #4
At the starting gun, a runner accelerates at 1.3 m/s2 for 5.4 s. The runner's acceleration is zero
for the rest of the race.
(a) What is the speed of the runner at t = 1.7 s?
(b) What is the speed of the runner at the end of the race? m/s
Student A: I am having trouble setting up this problem. I am looking through the book but
for some reason my head is drawing a blank, and helpful suggestions??
Dr. Man: Again, what are known in question (a)?
a=
v_0=
v_t=
t=
delta_x=
What is asked in question (a)?
v_t, right? what is t for question (a)?
What is not known and also not asked, so that not needed? Always choose the equation which
doesn't involve the item which is NOT needed.For (b), what is t? Also, start HW EARLIER!
It's not likely to get answers from me this late right before due time.
Topic: HW# 2a #5
A car is traveling due north at 23.0 m/s.
(a) Find the velocity of the car after 4.50 s if its acceleration is 1.60 m/s2 due north
(b) Find the velocity of the car after 4.50 s if its acceleration is instead 1.95 m/s2 due south.
Student A: I'm not sure which equation to use or how to do either of these, can anyone help?
Student B: you can use the equation:
v=Vo+a*t
since you know all those factors. and then make the acceleration negative value for part (b)
because it's going in an opposite direction.
Dr. Man: Make a table of the following:
a, v_0, x_0, t, v_f
which ones are know? What is asked in question (a)? v_f Which one is not known and
also not asked so that not needed? Always choose the equation that doesn't involve the one
not needed. Remember to define positive direction first. Since v_0 is due north, it make
sense to define north to be positive. Then I was reminded in part b you need to pay attention
to the sign and direction of a.
Topic: HW 2a #7
(a)Find the x and y components of a position vector, r, of magnitude r = 51 m, if its angle
relative to the +x axis is 40.0°.
(b) What if the relative angle is instead 65.0°?
Dr. Man: You are required to read Chapter 3 before Thur's class. This kind of HW is to
test whether you finish the reading assignments. You need to understand and use sin
and cos well for this Chapter. We will finish Chapter 3 in the next lecture.
Topic: HW 2b, #2
Bill steps off a 2.8 m high diving board and drops to the water below. At the same time,
Ted jumps upward with a speed of 4.2 m/s from a 1.0 m high diving board. Choosing the
origin to be at the water's surface, and upward to be the positive x direction, write the
x-versus-t of motion for both Bill and Ted. (Use g and t, as necessary.)
Student A: I thought i had the right equation and plugged in the numbers accordingly, but
my answer was wrong? Does anyone have any suggestions on how they answered the
question?
Student B: The entire question is unclear about what it's asking for. I understand that I'm
supposed to write my answer as if I'm plotting it on a graph, but I get punished for trying
answers and trying to figure out what the program wants. I can find an equation that would
serve as the function of x vs t, but the system doesn't seem to accept it.
A function where f(x) = something-with-t that would represent something that could be
graphed. It is also unclear as far as the significance of the symbols and I've noticed that
units are pointless. Is g positive or negative? There goes a try. Can we just use 4.9 if we're
unsure?
Student C: I got the first part right, but I don't seem to get the second part right. What I'm
trying to do is to make an equation for the part when he goes up and another equation for the
part when he falls down, and combine the two, but I don't understand how to do it.
Dr. Man: In this question they don't ask you the position at t=5s or t=0.3 s. It want you to show position x as a function of t.
The webassign is not as smart as a real human instructor.... :)
It doesn't reorganize your answer unless you follow its symbolic rules, which are not hard if you pay attention. Actually we practiced how to answer this kind of question in “Intro to webassign for phys.” Webassign has its shortcoming. But it is already very generous that you can try 5 times. If a human instructor is grading ur work, there is no way for you to try 5 times.
Please review question 10 in the first assignment “Intro to webassign for phys.” In that question, it shows you a right triangle. The two side lengths are a and b respectively. They ask you the length for the hypotenuse, c. You know that The square on the hypotenuse is equal to the sum of the squares on the other two sides.. So, c=(a2 +b2)1/2 ; Check what you need to input into the box to answer that question:
1, you should not include the sign “=”
2, you should not include the whole equation,
You should only put the right side of your answer in the box, : (a2 +b2)1/2
3, you need the right format: you can type:
sqrt(a^2+b^2)
you can also type: sqrt(a*a+b*b)
you can also type: (a^2+b^2)^0.5
Same thing for this question, do not put “=” in your answers.
Just put the right side of your function into your box:
For example, If you believe that Bill's position x at time t will be equal to 123 + t g + (1/2 * g^2 )- 5t
You just put 123 + t g +( 1/2 * g^2 ) - 5t in the box.
DO NOT PUT x=123 + t g + (1/2 * g^2) - 5t in the box. Notice that you should use g instead of 9.8 for answering this one.
You can clearly see the example answer above is not correct, just to show you the way of formatting
Also positive or negative signs are very important. When the answer is marked wrong,
something is wrong. Think more clearly and try more. Learn from mistakes.
Student C: Nevermind. I don't need to do it that way. Writing an equation for "going up"
actually accounts for the whole trip.
Dr. Man: Right. it is important to realize that one equation with gravity acceleration will
decide the whole motion after jumping up. You do not need to break it into two part of
motion.v= v0+ at and x=x0+v0 t+1/2 a t^2
are valid for any time after jumping. (before and after the maximum point) v0 and x0 are
fixed. a is also fixed. Pay attention to their signs.
Topic: HW 2b, #3
A hot air balloon is rising upward with a constant speed of 10 m/s. When the balloon is 2 m above the ground, the balloonist accidentally drops a compass over the side.
a.) How much time elapses before it hits the ground?
b.) What is its speed just before it hits the ground?
c.) How high above the ground is the balloon when the compass hits the ground?
Dr. Man: For part B:
Pay attention to the difference between velocity and speed.
For example: Velocity = V=V0+at and has directions. (positive or negative directions).
Speed is only the magnitude of velocity . (The size of velocity, the absolute value of velocity)
No matter you drive north or south, you speed is always >0
But your velocity can be >0 or <0 depending on the direction.
Part B asks speed, not velocity. You can find velocity and only answer its magnitude, not including the sign.
Topic: HW 2b #4
A diver springs upward with an initial speed of 8 m/s from a diving board that is 2 m
above the water. Find the total time the diver is airborne.
Student A: How do we account for the additionally 2m, since it's not symmetric
throughout the motion?
Dr. Man:
1, define positive direction.
2, identify which part of motion you are studying?
the total time the diver is airborne.
So, the initial was on diving board, the final position was on water.
Again what are known? v0 and a are straightforward, anything else known?
vf or displacement delta_X from the initial position to the final position?
Student B: I was thinking that I would calculate the time between the diving board and
the maximum using the initial velocity and the velocity at the top. Then I'd find the
velocity as the diver hits the water and use the velocity at the top to calculate the time in
the second part of the motion. Is my thinking correct?
Dr. Man: Remember what I said during the lecture,
the raising part and the falling part can be included in one equation. It is ONE same motion with CONSTANT acceleration, which never changed.
That's why in the sample question in the lecture, the first part from ground to ground, we could set one equation to solve v0, (knowing delta_x, t, and a)
Here similarly, raising and falling part is ONE same motion. (look at the v vs t gragh on lecture notes.), though v flipped direction.
As a result it is easy to set one equation to solve t (knowing delta_x, v0 and a). Just pay attention delta_x here is different from the sample question we solved in the class. What is delta_x for the WHOLE MOTION? from the board to water?
Attention delta_x is displacement, not total distance.
You can always verify your answer later by treating this motion into two different part. Find the time spent to raise and find the time spent to fall, and you will find they add together equal to the total t you get using the above method
Topic: HW 2 #6
A person on horseback moves according to the velocity-versus-time graph shown in Figure 2-32. (The vertical axis is marked in increments of 2 m/s and the horizontal axis is marked in increments of 4 s.) Find the displacement of the person for each of the segments A, B, and C.
Student A: I'm pretty sure I've been using the correct equation to find the displacement for
this problem. I used the same equation for all three segments but only got segmentB &
segmentC correct. However, for segmentA, it said "Your answer differs from the correc
answer by 10% to 100%."
Student B: I didn't use an equation for #6. Remember that the area under a velocity vs. time
graph is the displacement.
Student A: I was using the area formula. I got the same answer for segment A that
I did for Segemtn B. Is that wrong?
Dr. Man: Right. It is ok to use the area method.
It is also ok to use equations about delta_x.
Just be careful to identify what are directly known for that segment.
time during THAT SEGMENT is easy to figure out.
v0 and vf of THAT SEGMENT are also directly readable.
v (average) of each segment is also easy since v0 and vf are known.
Then you know which equation is the best for finding delta_x from the figure.
acceleration of each segment can also be calculated. But you need to calculate slope...
To use an equation involves acceleration can solve delta_x, but to use an equation
directly use v_ave or v0 and vf, but doesn't need acceleration can be faster.
Topic: HW 2b #7
A 27 pound meteorite struck a car, leaving a dent 18 cm deep in the trunk. If the meteorite
struck the car with a speed of 560 m/s, what was the magnitude of its deceleration,
assuming it to be constant?
Student A: I don't really know what equation to use for this. I tried solving for time
using (Time= Velocity * Distance), the distance being the 18cm dent in the truck and
then solving for acceleration but that didn't work. Any hints?
Dr. Man: First, there is no an equation of Time= Velocity * Distance.
Did you mean Time= Distance/Velocity? Please check the dimension of these two above
equation and realize why the first one is definitely wrong. Second, Time= Distance/Velocity