Exercise #2
Which of the following are vectors: (a) force, (b) temperature, (c) volume, (d) age?
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1. No. The distance traveled sets the limit on the magnitude of
the displacement.
2. Using vector notation, we have the displacement at d=6i+7j-4i-2j=2i+5j. d is a vector. i and j are the unit vectors. By Pythagorean theorem, the displacement should be 5.4 km, less than 7 km.
3. (a). Temperature can be negative or positive but it does not have any direction.
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1
Tue Sep 9 10:46:45 2008
No it can't. Your distance traveled can be greater than the magnitude of
displacement, but never can it be the opposite. If you move right 5m and
then left 2m, your distance traveled is 7m but your displacement is only 3m.
So the magnitude of a particle's displacement can never be greater than the
distance traveled.
Neither, you are exactly 7km away, because you end up being 5km north and
2km east of where you originally started.
b
2
Tue Sep 9 11:04:03 2008
No, because displacement= x(final)- x(initial); therefore, it will always be
the same or less than the total distance traveled. Also, if looking at the
concept from the point of a right triangle, the displacement (hypoteneuse)
is usually less than the total distance traveled.
The car's desplacement is less than 7 km, because after cancelling out the
distance going backwards, the car travels 2 km east and 5 km north. The
vector would be less than seven km.
a
3
Tue Sep 9 12:26:19 2008
No, the distance traveled will always be equal to or greater than the
displacement. For example, if traveling 5 units left then 8 units right, the
total distance traveled would be 13 whereas the the displacement was 3
units.
Less than 7. Displacement is defined final position - initial position, and
a displacement vector represents this number. Therefore, the displacement of
problem #2 would be 4.47 ( the square root of 20).
a
I think we should have more than one days notice to complete warm-ups
because some days I as very busy.
4
Tue Sep 9 12:46:23 2008
no, but the distance traveled can be greater than the magnitude of
displacement. Backtracking on a path might increase distance traveled while
decreasing the particle's displacement, the reverse relationship does not
exist.
less than, because in the end the particle is closer than 7 km away.
d
i found the above assignment a little difficult, because I have a very
limited background dealing with vectors and displacement; but the idea seems
interesting.
5
Tue Sep 9 13:13:01 2008
No. The two measures can be equal and the distance can be larger than the
magnitude but not vice versa. This is because the magnitude is a+b=s where s
cannot be larger than a+b.
It is less than 7 km because at the end of the travel the car is only
actually only 5 miles north and 2 miles easy of where it began. This makes
its displacement the square root of 29.
a
6
Tue Sep 9 13:34:18 2008
No. Because displacement is the straight line distance between a starting
poing and a resting point, it is by definition the shortest route between
the two points.
Less than seven. The final resting point of the car puts it at 2km east and
5km north of its original position. Using pathygorean theorem, the magnitude
of the displacement is roughly 5.3.
a
7
Tue Sep 9 13:53:29 2008
No, the a displacement is always a straight line and the shortest distance
between to points is a straight line so it would be imposable to travel less
than what your displacement.
Less, even though the car traveled a total of 19km the displacement is
around 5km due to the fact that it backtracked by traveling east then west
and north then south.
a
8
Tue Sep 9 13:54:32 2008
No, the displacement vector gives you the straight line (shortest distance)
from point A to point B. Displacement vectors represents the overall outcome
of a motion, not the motion itself.
Less than 7 km, because the displacement shows you the shortest distance
from the head of the 6 km east to the tail of the 2 km south.
a
9
Tue Sep 9 14:38:27 2008
No, displacement is the "net" movement, so therefore, it can be less than
the total distance traveled but not more.
Less than 7, because displacement isnt concerned with the actual distance
covered in a trip, jut the origin and final positions. By using the
pithagorean theorem, I figured out the distance between final and original
position to be just over 5.
a
10
Tue Sep 9 14:58:49 2008
No, the magnitude of a particle's displacement will always be less than or
equal to the total distance traveled. When a particle takes a different path
to get from point A to point B, the magnitude will be less than the total
distance traveled.
Less than 7 km, if you transfer the car's displacement onto an x,y
coordinate plane, you can find that if the car's initial position is at
(0,0), it's final coordinate is at (2,5). This gives a displacement of 2.5
km, far less than 7 km.
b
11
Tue Sep 9 15:26:24 2008
No. The magnitude of displacement cannot be greater than the distance
travelled because magnitude of displacement, which has no sign + or -, is
simply the difference from point b to point a, whereas the distance traveled
may or may not be a straight line from point a to b. The distance is each
unit that the particle travels, but the magnitude of displacement is the
difference from one point to another.
The displacement is less than 7 km because the displacement of the car is
actually only 2 km to the east (6 east - 4 west) and 5 km to the north(7
north - 2 south). By pythagorean theorem, a^2 + b^2 = c^2, so 2^2 + 5^2 =
29. So the distance actually travelled is the square root of 29 which is
less than 7 km.
a
It was not overly difficult, but it made me think about the definition of
vectors, displacement, and distance.
12
Tue Sep 9 15:37:06 2008
No, a particle's displacement is the final position of the particle minus
the initial position of the particle, regardless of its path. The distance
traveled, however, depends on the path. Therefore, a particle's displacement
is the difference from final position to initial position, and this will
always be less than or equal to the distance traveled.
It will be less than 7 km because collectively, the car traveled 2 km east
and 5 km north. If one uses vectors to solve for the vector sum, it will be
the hypoteneuse of the right triangle created by moving 2 km east and 5 km
north. The value of this hypotenuse will be less than 7 km.
a
13
Tue Sep 9 15:47:26 2008
No, because the magnitude of displacement does not give any indication of
the path traveled by the particle from point A to ending point B.
Less than 7km because you traveled "back" along the 6km east when you went
4km west; the same can be said for the north and south direction.
a
I'm not sure I quite understand the concept of adding vectors, especially in
cases like problem 2. I think this would be a good section to spend a little
extra time on in class or in SI sessions.
14
Tue Sep 9 16:19:56 2008
No the magnitude of a particle's displacement must be equal or less than the
distance traveled. Displacement is a particle's endpoint minus the
particle's starting point. If the particle went 10 meters forward and 5
backwards the displacement would only be 5 m but the distance traveled would
be 15 m.
The car's displacement is about 5.4 km NE of the starting point. The
displacement is less than 7 because even though the car drove a long
distance its vectors subtracted from one another. Overall it moved 2 km east
and 5 km north, but displacement is a straight line from the endpoint to the
starting point which, in this case, is about 5.4 km.
a
15
Tue Sep 9 16:29:19 2008
No, even if the particle went in a straight line, the maximum displacement
would be the distance traveled.
The displacement is less than 7 km because the car ends up 2 km east and 5
km north of its start point, so the distance would be the square root of
29(5 squared + 2 squared).
a
16
1. No, the distance travel will always be equal to or greater than the
magnitude of a particle's displacement. The distance traveled includes the
bends and curves that you take on your way to your destination. The
magnitude of a particle's displacement is a straight line from point a to
point b. If you walked in a straight line from a to b then your two quanties
will equal one another.
2. It would be less than 7km, it would be the square root of 29. To figure
it out you draw the distances, and get a square type diagram. The
displacement would be the end point to the starting point. I drew a right
triangle with the remaining open space and use the pythagorean theorem to
figure out the distance between point a and point b.
3. a
17
Wed Sep 10 12:11:09 2008
No, because distance traveled is always positive and you cannot have
negative distance. However, you can have negative displacement which
prohibits displacement to never be greater than distance traveled.
Yes it is less than 7km. I believe the displacement is the square root of
29. Allow that displacement is concerned is the starting and ending
location. the total distance traveled is 19km but the displacement is less
than 7km.
c
18
Wed Sep 10 12:44:12 2008
No because if the particle's displacement is from point a to point b, the
distance will eventually get to point b to point a but may be forced to go a
different route and lengthen it. Even if the distance traveled is in a
straight line from point a and point b exactly along the path of the
displacement, it is equal, never less than the particle's displacement.
Its displacement is less than 7 km. If you figure the displacement as a
right triangle, the two legs of the triangle would be 2 km and 5 km, making
the length of the displacement the square root of 29 (less than 7).
a
19
Wed Sep 10 12:44:14 2008
No, because the magnitude is the sum distance between a point (a) and (b)
hence it does not take in account for the total distance traveled which
could have included negative direction or any weaves.
It's displacement is less than 7km because if you draw a graph and calculate
the lengths of the x and y you can use the formula for a right triangle to
calculate the length of the hypotenuse, which is the displacement to be less
than 7km.
a
20
Wed Sep 10 15:11:07 2008
No. The magnitude of a particle is the same as the displacement of the
particle.
Less than 7 km.
The actual displacement as calculated by the distance formula is:
(3^2+5^2)^(1/2)=34^(1/2)<7
a
21
Wed Sep 10 16:28:14 2008
No a particle's displacement can never be greater than the distance
traveled. This is because displacement is the change in position of the
particle, which is controlled by the distance traveled, so at most the
displacement could equal the distance traveled. However, the total distance
traveled could be greater than the displacement because displacement doesn't
take into account the path taken to get to it's final position.
Its displacement is less than 7 km because the car changed directions back
on itself lowering it's position change. Originally, the car was heading
north and east, but then it went west and south. The magnitudes of the
vectors west and south were enough to lower the displacement to a value less
then 7 km.
a
22
Wed Sep 10 16:37:16 2008
No the displacement can never exceed the total distance traveled because the
displacement is the distance from the origin while the distance traveled is
the overall distance of the particles path whether forwards or backwards.
Less because the car ends up 2 km east and 5 km north which is less than 7
km from the origin.
a
23
Wed Sep 10 16:42:31 2008
No.If all the displacements are positive, they will only be equal to the
distance traveled. If some displacements are negative, then the displacement
will only be shorter than the distance.
No. The net displcement of the car is 5km north and 2km east. Then the total
displacement is square root of 4+25, which is less than 7km.
a
24
Wed Sep 10 20:43:47 2008
No. A particle's displacement is defined by a change in position. If a
particle changes position from its origin and goes a certain distance, say
30meters, but then returns to its origin, the displacement is zero but the
particle traveled a distance of 30 meters.
The displacement is less than 7 km. If the car continues on its path, more
south, it will come close to its origin. At its resting point as described
above the car is 7km from its origin.
a
The second question confused me a little.
25
Wed Sep 10 21:13:54 2008
no it cant displacement is how far an object has acually moved from its
starting point this can only be equal or less than the distance traveled.
less than because a straight line between the starting point and ending
point would be less than 7km
a
26
Wed Sep 10 21:34:38 2008
No, magnitude of displacement will only tell the overall effect of the
movement, not the actual movement itself. the actual movement can be equal
to the displacement but never less than the magnitude of the displacement.
The displacement would be less than 7. It would actually be square root of
29 and this is due to the fact that the car nearly came back to its origin
during the course of travel. Displacement is the distance between the origin
and final spot.
a
Great excercise...made me think pretty good and still enjoyable.
27
Wed Sep 10 22:07:48 2008
The straight line connecting two vectors (the vector sum) is a shorter
distance than the distance covered by its component vectors.
The displacement is less than 7km (about 5.4km) because it only involves the
difference between the initial and final positions of the car. The overall
distance covered during the trip is 19km, the sum of all four distances.
a
Are the warm-ups always due on Fridays?
28
Wed Sep 10 22:27:45 2008
no because it cannot be farther from the start than what the object
traveled.
It is the square root of 4 squared plus 5 squared, the square root of 41
which is less than 7. I found this out by setting up a triangle and solving
for the hypotenuse.
a
29
Wed Sep 10 23:04:31 2008
No because the magnitude of a particle's displacement is a direct vector and
the distance traveled can be on the same line of curved thud being bigger
then the magnitude.
Less then 7 becasue the magnitude of the resulting vector is the square root
of 29 which is less then 7.
a
30
Wed Sep 10 23:18:21 2008
No, because even if its a straight line, the magnitude can only be equivlant
to the total distance. (Not greater than)
The displacement is less than 7 km because the magnitude of the vector is
the square root of 29 which is less than 7.
a
31
Wed Sep 10 23:36:25 2008
No. Vectors can be added, but even then the vector sum is the net
displacement, which will not be more than the total distance traveled.
Overall, the car ends up going 2km east and 5 km north. Addition of these
vectors results in a number larger than 7.
a
32
Thu Sep 11 09:07:38 2008
Yes, if, for example, you traveled a distance a, 5 m due north and a
distance b, 6 m due east, the two vectors will meet at a 90 degree angle,
with the total displacement being the hypotenuse of the right triangle,
which is also the longest side of the triangle.
The displacement is less than 7 km, it is the square root of 29 m, because
the car has been displaced 5 km north and 2 km east. The car's total
displacement is equal to (5^2 + 2^2)^(1/2) or the square root of 29, which
is less than 7 km.
a
I thought number two was confusing even after looking at examples within
chapter 3. I would like further explaination on this example. I thought the
previous chapter's warm-up was more interesting. Is there any way the
solutions to these warm-ups could be posted online Friday after class?
33
Thu Sep 11 17:14:43 2008
No. The maximum magnitude of a particle's displacement would be the distance
traveled because the particle would travel in a straight line.
The displacement is less than 7 km because the final resting place is 5 km
north, 2 km east which is less than 7 km from the starting point. One way to
verify this is to view the two vectors in terms of a right triangle and use
the pythagorean theorem. 5*5 + 2*2 = 29 which is less than 7*7=49
a
34
Thu Sep 11 17:26:08 2008
While magnitiude is always positive, distance travelled can be negative.
Therefore, a the magnitude (absulute distance) of a particles distance can
be greater than the distance travelled.
The displacement is less than seven. Displacement is defined as the absulte
distance between the initial and final points. In this case, while the
distance travelled is greater than 7km, the distance between the starting
and final locations remains less than 7km.
a
35
Thu Sep 11 18:07:44 2008
No, the magnitude of a particle's displacement cannot be greater than the
distance traveled. Distance can be greater than displacement but
displacement cannot be greater than distance because of the vector.
Less than 7km. Because displacement is the change in position. Since the car
traveled approximately 5 miles back towards the beginning point the
displacement is less than 7km.
a
36
Thu Sep 11 18:09:46 2008
No. The distance travelled refers to the total distance. A particle's
displacement refers to the total change in position. If a car goes 20m in
the positive direct and 15m in the negative direction, its distance traveled
is 35m but its displacement is only 5m. Thus, a particles displacemet can
only match that of the distance traveled and can never exceed that
magnitude.
The total displacement must be less than 7km because 7km is the largest
magnitude of the 5 listed positions. The only way the displacement could
match or exceed 7km is if the total distance for one direction (north,
south, east, or west) had a sum of 7km+
a
37
Thu Sep 11 18:32:41 2008
No. If the particle goes in a straight line the distance will be equal to
the magnitude of displacement. If the particle changes direction then the
magnitude will be smaller and the distance will be farther.
Less than 7. From starting point to ending point ends up being only 2km east
and 5km north and the square root of 29 is less than 7.
a
38
Thu Sep 11 19:16:26 2008
No; the magnitude of a particle's displacement is the distance between the
original and final positions. In other words the magnitude is the distance,
but without a sign.
The over all displacement would be greater than 7km because the car is
traveling a total of 19km even though it is back tracking its path by going
west and south.
b
the second question was kind of confusing because I wasn't sure if since the
car traveled both east and west, whether or not they were able to cancel one
another out.
39
Thu Sep 11 20:07:03 2008
No. The particle can meander all over the place but it's magnitude is the
net result of its position. This net result could equal but never be greater
that the particle's displacement.
Less than 7km. 7km is the farthest the car travels in one direction from its
initial position. After the subsequent directions traveled. The particle is
closer to its initial position than 7km.
a
40
Thu Sep 11 20:43:12 2008
No, the distance traveled will either be equal to or greater than the
particles displacement because the displacement will be the vector sum of
the distance traveled, and thus can never be smaller.
It's displacement is less than 7 km, because the final destination is 2
miles east, 5 miles south of it's origin and its hypotenuse is the square
root of 29, which is less than 7 km.
a
41
Thu Sep 11 21:19:58 2008
No. when adding the two distances up they will always be larger than the
particle's displacement.
no. the new coordinate is 5,2 meaning that the distance from the origin is
5.2 which is less than 7
a
42
Thu Sep 11 21:24:19 2008
Yes if you were to move in the opposite direction.
Less becasue when you went west your direction changed which changed the
Displacement.
a
43
Thu Sep 11 21:50:20 2008
No; magnitude is directly correlated to displacement, which could end up
being a number less than distance traveled, but never more. For instance: if
an object moves from x=5 to x=200 and back to x=10, the magnitude (which is
the absolute value of displacement) would be 5, but distance traveled is
much more than that.
Less than 7; Displacement only depends on initial and final position. In
this example, after all of the traveling, the car is only 6 km away from the
initial point.
a
44
Thu Sep 11 21:51:22 2008
Yes, the magnitude of a particle's displacement could be greater than the
distance traveled. This can be explained with the use of a simple example.
Lets say I start traveling towars Yankton but half-way I realize I forgot my
wallet. So I turn around back to Vermillion, grab my wallet and head back to
Yankton. A trip that says I am 32 miles away from where I started, took me
64 miles (the magnitude).
The displacement for this car is less than 7 km. The displacement of an
object is just the change in position (from original position to final, not
the path taken). This question can be analyzed using simple geometry and the
use of triangles. The overal displacement laterally is 2 km east (6 east
minus 4 km west equals 2)and the overal displacement vertically is 5 km
north (7km north minus 2km south equals 5). This gives us a triangle with
sides 2 and 5. Using the pathagreium thereom, we find the hypothenous to
have a vector of magnitude of 5. 38, which is less 7 km.
a
45
Thu Sep 11 21:54:11 2008
No, because the displacement can only be the same or smaller than the
distance traveled. The distance traveled cannot be negative as it does not
have direction, so the displacement magnitude can't be larger.
Less than 7km, because the displacement on the x-axis is 2 km and the
displacement on the y-axis is 5km, so the total displacement is a little
over 5 km.
a
46
Thu Sep 11 21:54:28 2008
No, the magnitude of displacement will always be equal to or less than the
distance traveled, because a particle can't be displaced by more than the
distance that it traveled.
The displacement would be less than 7 km because the end point is only 2 km
in the x direction and 5 km in the y direction from the starting point.
a
47
Thu Sep 11 22:00:57 2008
No, the magnitude is the absolute value of the displacement. The direction
may change but the distance traveled will remain the same.
The displacement is equal to 7km. 6kmE-4kmW=2kmE and 7kmN-2kmS=5km
a
48
Thu Sep 11 22:08:15 2008
The magnitude of a particle's displacement can not be greater than the
distance traveled. Displacement is the measure from point A to point B, and
if point A and point B are in the same location, displace would be zero
(even if there were other stops away from this point along the way). Where
as the distance traveled in that example would be the distance from point A
plus all the other stops along the way before getting back to point B. So,
while distance can be greater than displacement, displacement cannot be
greater than distance.
The displacement is less than 7km because if these directions were plotted
on a graph, the final end point of destination would be at a coordinate
point (2,5) and the start location at (0,0). To find the displacement one
would need to find the slope of the line from point A to point B. Which
would mean (5-0)/(2-0)=2.5km which is less than 7. However, the total
distance traveled would be greater than 7 km.
a
49
Thu Sep 11 22:25:30 2008
The displacement can never be greater than the distance traveled because it
only involves the start and end points. Displacement will always be greater
or equal to distance travelled.
The displacement is less than 7 because the car travels in all for
directions which eventually starts to bring it back towards the start point.
a
the difference between vectors and scalars is confusing
50
Thu Sep 11 22:43:16 2008
No, the distance traveled will always be greater than or equal to the
particle's displacement.
The displacement is less than 7km. The end position would be 2km east (6km
east-4km west)and 5km north (7km north-2km south) of the start position.
Using the pythagorean theorum, the displacement is 5.4 km.
a
51
Thu Sep 11 22:55:55 2008
No because the magnitude is the directional vector of force on the object
greater than 7 because the legs of a 90 degree triangle can not be added
together and equal the hypotenuse of the triangle
a
52
Fri Sep 12 00:21:27 2008
No, because magnitude is represented as a straight line, the shortest
distance between two points, and the actual path could have not been a
straight path. So, the actual distance traveled could be greater than the
magnitude but not the other way around.
It is equal to seven because the car travels 6 km east but then ends up
going 4 km west, making the horizontal displacement 2 km. The car also
travels 7 km north and then 2 km south, making the vertical displacement 5
km. The total displacement would then be 7 km.
a
53
Fri Sep 12 00:36:23 2008
No, the magnitude of a particle's displacement cannot be greater than the
distance traveled. The distance traveled includes the sum of all lengths
traveled, in any direction, in getting from one place to another.
Displacement, on the other hand is just a single straight line connecting
the initial position to the final position. And the shortest distance
between two points is a straight line. Therefore, the magnitude of a
particle's displacement can possibly be equal to, but never greater than the
distance traveled.
The displacement of the car is less than 7km. At the end of the car's
traveling, it is located 2km east and 5km north. So, consider these values
as the legs of a triangle. The hypotenuse of the triangle is the
displacement. The sum of the legs is 7km. The length of the hypotenuse
cannot be greater than the sum of the two legs. The shortest distance
between two points is a straight line. Therefore, the displacement must be
less than 7km.
a
54
Fri Sep 12 00:36:50 2008
The value for magnitude could be greater than the distance traveled because
magnitude doesn't include direction.
the displacement is equal to 7 because 4 and 2 are negative directions.
a
Question one's answer isn't clear to me. Wouldn't the magnitude of
displacement equal the total distance of a traveling particle? If a particle
moves south than north than the vectors would cancel out but they would
still travel a certain distance, their displacement would just equal zero.
Could we go over this in class?
55
Fri Sep 12 00:44:36 2008
No because the amagnitude takes into consideration just the distance from
the origional starting point to the end point. The distance traveled must
always be equal to the magnitude of displacement if not more.
No it is 5.385km from its origional starting distance. So it is less than 7
km from its starting distance.
a
Not really sure about my answers but I remember vectors from highschool
physics and trig.
56
Fri Sep 12 00:58:34 2008
Yes,because if the displacement is negative the magnitude would be larger,
because it would be positive.
The car's displacement would be equal to 7 because the car travels 6 km but
comes back 4, which is 2, and then goes 7 but comes back 2, which is 5,and
gives you a total of 7
a
I was a bit confused with the second question. I didn't know whether I
should figure it the way I did or in a straight line.
57
Fri Sep 12 01:19:38 2008
No, Because the magnitude is the total distance from the starting point.
However, the distance traveled can be greater than the magnitude.
Less, Because the displacement is less than 7 even though the distance is
greater. The distance from the starting point is less than 7.
a
58
Fri Sep 12 02:41:42 2008
No, the magnitude of a particle's displacement can never be greater than the
distance traveled. This is because displacement only ever represent the
overall effect of motion, not the motion itself. Displacement is always
meant to be the shortest distance between two points, and thus, it is
impossible for distance traveled to ever be less than displacement (or for
displacement to be greater than distance traveled).
The total displacement is less than 7. This is because displacement only
considers the net distance away from the origin. In this problem, the car's
net distance north is only 5 km (it originally goes 7 km north, but then
goes 2 km south)and the net distance to the east is only 2 km (it originally
travels 6 km to the east, but then goes 4 km to the west). If we form a
vector by making a triangle from the 5 km north and 2 km east vectors, that
vector (the hypotenuse) is equal to about 5.3 km, which is less than 7.
a
59
Fri Sep 12 07:26:12 2008
No, the distance could never be less than the magnitude of the displacement
vector because the shortest distance between two points is a straight line.
Equal to 7. A positive 6 plus a positive 7 minus a negative 6 due to
backtracking equals 7.
a
60
Fri Sep 12 07:37:55 2008
No. The displacement can be less than the distance traveled, if part of that
traveling involved moving in a negative direction. But displacement will
always be at least equal to, if not less than the total distance, because
you cannot travel farther than you traveled.
The displacement is less that 7km because of the negative movements (4km
west and 2km south are in the opposite directions of the 6km east and 7km
north). The total distance away from the origin is found through the
pythagorean theorem, a^2+b^2=c^2.With the negative movements, the final
point ends up being 5km north and 2km east. If this is graphically
represented, 5 and 2 are two of the legs of a right triangle, with the
hypotenuse (c) equaling the displacement. Therefore, c=sq.rt.(5^2+2^2),
making c~5.4, which is less than 7
a
61
Fri Sep 12 07:38:08 2008
Yes. a simple example demonstrating this concept could be looking at my walk
on campus. Let's say i am walking from McDonalds to Akeley Science room 125,
except on my way to the class i decide i want to go back and get another
McChicken. The trip says that i am 1 block away, but in reality i walked
almost twice as far.
Its displacement is less than 7 km. The reason is that although it looks as
though the car traveled farther than 7 km, the total distance is not used in
displacement; rather, the vector sum is used. Which is a single displacement
line drawn from the starting point to the end point of the traveled
distance.
a
62
Fri Sep 12 07:43:10 2008
Yes, because if you were to travel toward an object and then suddenly turn
back half way on ur trip to that object and return to the start to grab
something and then continue once again toard the object, your displacement
would be as far as you had originally traveled from start to the tuirning
around point, but your magnitude would be all the distance traveled added up
and be far greater.
Less thasn 7km. This is so because displacement reads change in position,
start to finish, not total distance. Using the pathagriam theorem we can
figure that one side is 2km, the other side is 5km. So, the other side is
then 5.38, which is less than 7km.
a