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Kinematics Problem

Three particles A,B,C are situated at vertices of a equilateral triangle ABC of side ‘a’ m at t=0 Each of the particle moves with constant speed ‘ v ‘. A always has it’s velocity along AB ,B always has it’s velocity along BC , C always has it’s velocity along CA. Derive the equation of trajectory of any one particle (means find y=f(x) i.e. relation between y and x coordinates of the particle). Also find the rate of rotation of the triangle formed by joining the lines connecting the three points as a function of time. If you think it requires more information such as acceleration etc., introduce them if needed.
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Numerical Problems from Kinematics

1.      A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

2.      Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

3.      A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

4.      A ball thrown up vertically returns to the thrower after 6 s.
Find
(a)     The velocity with which it was thrown up,
(b)     The maximum height it reaches, and
(c)     Its position after 4 s.

Size of body neglected in Circular motion

Why in  circular  motion the  size  of  the moving body is  neglected?

Answer:

Not only in circular motion, but in entire kinematics we discuss the case of point objects only. In kinematics we study motion only and not the objects undergoing the motion.

A point object is an object having negligible dimensions compared to the distance travelled by it. For example, a car moving from Delhi to Mumbai can be considered a point object.

Further, when we discuss the concepts in Kinematics, even a large object is dealt considering the entire body to be a single point. This point will be the centre of mass of the body.

If a person is revolving a stone tied to a string, the centre of mass of the stone is taken for determining the length starting from the point about which it revolves.

In dynamics, we have to consider the shape and size of the body also.

Kinematics – Signs of displacement and Velocity

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

[This question was asked via our mobile interface available at http://m.askphysics.com (mobile users visiting www.askphysics.com will be automatically redirected to the mobile version). We are happy that many of our visitors are using the mobile version and utilizing the facilities]

Answer:

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.

Problem from Kinematics

the distance between Ahmadabad and vadodara is 100 km.two trains set off simultaneously from Ahmadabad towards vadodara each other.the speed of these trains are 45kmh-1 And 30kmh-1.when will they cross each other?

Problem from Kinematics

A particle moving with constant acceleration describes in the last second of its motion 9/25th of the total.if it starts from rest how long is the particle in motion and through what distance does it move if it describes 6 cm in first sec

Problem from Kinematics

The speed of a train changes from36km/h to 72km/h in 10sec. calculate the distance travelled during this time

Answer:

u=36km/h = 10 m/s

v=72 km/h=20 m/s

t=10 s

S=?

use v = u+ at to find a

a = 1 m/s2

Use v2-u2=2aS to find S

S = (400-100)/2 = 150 m

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