Under what condition will the magnitude of displacement be equal to the distance travelled by an object?
When an object moves in a straight line without coming back, it's displacement and distance magnitude will always be equal.1)why not travelling backwards?Displacement is is a vector quantity and distance is a scalar quality.So if you walk on a line ABC, and you walk from A to C and then walk backwards to B:Your total displacement will be, AC - CB this is because displacement is a vector and it considers direction.ButYou total distance travelled will be, AC + CB this is because distance is a scalar and it doesn't consider direction.Now it's clear that why the obeject shouldn't come back for them to be equal.2)Now, why a straight line?Look at the following common example:You can easily understand if the object doesn't follow a straight line its distance is larger than its displacement.
How do you find the sum of four displacement vectors and with this find its magnitude and directional angle?
Question: The magnitudes of the four displacement vectors shown in the drawing are A = 11.0 m, B = 11.0 m, C = 12.0 m, and D = 27.0 m. Determine the magnitude and directional angle for the resultant that occurs when these vectors are added together. http://www.webassign.net/CJ/p1-42.gif The image is here, but if it doesn't show up for some reason, it has an x, y coordinate plane and gives the following information not stated in the question: A is 20 degrees N of W. B is true North. C is 35 degrees S of W. D is 50 degrees S of E. Thanks so much in advance!! I had a physics test today on this sort of information, and I was unable to answer a very similar question. There is an extra credit worksheet on our class forums online that is due tonight by midnight. I want to understand this and be able to get those extra points to make up for the botched test.
In the vector sum A + B = C, Vector A has a magnitude of 12.0 m and is angled 40.0 degrees counterclockwise..?
One way to do this is to figure out the x and y components of your polar coordinate vectors. A = 12m, 40deg, so Ax = 12 * cos(40) and Ay = 12 * sin(40) C = 15m, 200deg, so Bx = 15 * cos(200) and By = 15 * sin(200) (note both angles have to be from the +X axis) so in Cartesian coordinates: B = C - A = (Bx, By) Bx = 15 * cos(200) - 12 * cos(40) By = 15 * sin(200) - 12 * sin(40) The magnitude would be the length: sqrt( Bx^2 + By^2 ) The angle would be the: atan( By / Bx ), but you have to make sure you wind up in the correct quadrant since the tangent function has a period of 180 degrees, but drawing a little picture will help you figure out that that B has to be somewhat clockwise from -X - or by inspection of the signs of Bx and By.
Determine the magnitude and direction of the displacement if a man walks 32.5 km 45° north of east,?
the east component of the first segment = 32.5 cos 45 = 23km the north component of the first segment = 32.5 sin 45 = 23km the total east component = 38km the total north component = 23 km magnitude of displacement = Sqrt[38^2+23^2] = 44.4km the angle of the displacement is given by tan(theta) = 23/38 theta = 31.2 deg north of east
Two forces of magnitude 10N & 8N are acting at a point. If the angle b/w the two forces is 60°, how do I determine the magnitude of the resultant force?
Since typing all those formulas were getting cumbersome I just solved it on a piece of paper and uploaded it.For reference,I have taken vector A=10N, vector B=8N, the angle between vectors A & B asθ (theta) = 60 degrees.The resultant vector is C and at an angle ø (phi) with vector A.Notice another thing. I have taken ø between vector A and C. If I need to know the angle between vector B and C, say angle α (alpha)or just α = 60 degrees - θHope that helps.
If the magnitude of vectors A, B and C are 3, 4 and 5 units respectively and A+B=C, what is the angle between vector A and B?
It'll be a right angled triangle with the sides of 3, 4 and 5 as it is the triplet.First condition is given angle A = angle B + angle C. It implies that triangle is isosceles triangle.By both of the above statements the triangle is right angled triangle at A having angles 45°, 45°, 90°.*** हरि ॐ ***
Force and Vectors?
You will need to supply a drawing or at least a better description on the position of each charge. Let's assume that the charges are arranged as follows A B C D We compute for the resultant vector using x- and y-components: X: -B + Dcos(45°) = -3.10 + 1.00 cos(45°) = -3.10 + 0.707 = -2.393 N or 2.393 N repulsive force to the left Y: -C + Dsin(45°) = -6.80 - 1.00 sin(45°) = -6.80 - 0.707 = -7.507 N or 7.507 N attractive force downwards Computing for the final resultant vector Magnitude of force = √(2.393^2 + 7.507^2) = 7.88 N Direction of force: tan (θ) = 7.507/2.393 θ = 72.3° clockwise from positive x-axis
If vectors OA,OB, and OC have the same magnitude R and the angle between OC and OA and OC and OB is 45, what is the sum of these vectors?
Then angle between vector OB and OA will be 90° now break vector OC into its rectangular components..Like vector OCcos45° along vector OA and OAsin45° along vector OB then add them simple algebraically where |OA|=|OB|=|OC|=R such asR+Rcos45° along vector OA(1+1/√2)RAndR +Rsin45° along vector OB(1+1/√2)RNow apply formula resultant =(A^2 + B^2 + 2ABcos$)^(1/2) where $ is angle between themHere $ = 90°So we have sum =(√2 +1)R
A sailboat race course consists of four legs, defined by the displacement vectors A, B, C, and D as the drawing indicates.?
A sailboat race course consists of four legs, defined by the displacement vectors A, B, C and D , as the drawing indicates. The magnitudes of the first three vectors are A = 3.00 km, B = 4.60 km, and C = 5.30 km. The finish line of the course coincides with the starting line. Using the data in the drawing, find (a) the distance of the fourth leg and (b) the angle .
The vector sum of three vectors gives zero resultant. What can the orientation of the vectors be?
Let me describe you with an example . Let the vectors be forces Fa= 4N. Fb= 3N. Fc= 6NThen here the resultant of these vectors will be zero and forms the triangle(ie. These vectors represents the sides of triangle) when the sum of small vectors among these 3 vector ( here, 4N and 3N) will be more then the third one(ie greater vector ,here 6N)For ease,Smallest vectors above is 4N and 3NSum of these smallest vector = 7N which is more than the 6N( remaining another vector).. So it forms a ∆ and its resultant is zero . and its acceleration is also zero....