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# Category Archives: Vectors

## Displacement Problem

A particle has a displacement of 12 m towards east and 5 m towards north and then 6 m vertically up. Find the magnitude of the resultant of these displacements.

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The resultant of the first two displacements – 12 m and 5m in mutually perpendicular directions is

$\sqrt{\left&space;(&space;12^{2}+5^{2}&space;\right&space;)}=13$ m

The resultant of 13m and 6m in mutually perpendicular directions is

$\sqrt{\left&space;(&space;13^{2}+6^{2}&space;\right&space;)}=14.32&space;m$

## Kinematics – Signs of displacement and Velocity

When will displacement and velocity be negative ? Can you explain me about angular acceleration and angular velocity.

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The signs are conventional. We take one direction as positive and the direction opposite to it as negative.

When we discuss one-dimensional motion, it is easy to take the help of the number-line, whatever towards right is positive and towards left is negative. If the initial direction of motion is taken as positive and the further displacements does not make the body to go beyond the initial position in opposite direction, the displacement will be positive.

## Displacement – A Numerical Problem

A fly is sitting in the middle of a wall of a cubic room of side ’a’.If it flies and sit to one of opposite corner of the wall. Find its displacement?

If the fly sit on the same wall on the horizontal, (from A to B in the diagram) the displacement is “a
If the fly sits on the diagonally opposite corner on the same wall, (from A to C in the diagram) the displacement is “√2 a
If the fly sits on the corner on the longest diagonal of the room,  (from A to H in the diagram) the displacement is “√3 a

## Problem from Vectors and Kinematics

“Over the course of about six weeks in 1992, Aki Matusushima, from Japan, rode a unicycle more than 3000 mi across the United States. Suppose Matsushima is riding through a city. If he travels 250.0 m east oone street, then turns counterclockwise through a 120 degree angle and proceed 125.0 m northwest along a diagonal street, what is his net displacement?”

(The question will be answered soon. Visitors can send their response and comment to this post)

## Vector resolution and addition

Sir please send me the answers with the reasons and steps

1. the drawing on the left shows two vectors A and B, and the drawing on the right shows their components.

The scalar components of these vectors are:

When the vectors A and B are added, the resultant vector is R, so that RAB. What are the x– and y-components of R?
2. the displacement vectors A and B, when added together, give the resultant vector R, so that RAB. Use the data in the drawing to find the magnitude R of the resultant vector and the angle q that it makes with the +x axis.

In the first question as you can see that the X components (Ax and Bx being equal and opposite will cancel each other and therefore the X com ponent of the resultant is zero.

The Y components will add up and therefore the Y component of resultant = 3.4+3.4 = 6.8 m and this itself is the magnitude of the resultant as the X components have canceled out.

Hnit for Answer to the second question

## Resolution of Vectors

From: Vakas

Subject: Resolution with geometry
Message Body:
Dear sir,
I m not able to understand that when we resolute mg in a inclined plane that plane having angle theta,how does the vertical component goes through inclined surface & cos comp opp to N.
explain with geometry so that i don’t get confused in future in determining the two comps is right direction.

Ans:

In a right angled triangle,

$cos\theta =\frac{adjacent side}{hypotenuse}$

and

$sin\theta =\frac{opposite side}{hypotenuse}$

When mg is resolved into mutually perpendicular directions, one parallel to plane and the other perpendicular to the plane (opposite to the Normal reaction), the component which is adjacent to θ is the cos component and the one away from θ is the sine component.

Students find difficulty ,mainly in identifying the angle .

See the diagram below and try to analyze

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