114 - Kinematics Questions Answers

a particle is moving along a straight line such that its position varies with time t as x=6α [ t+ αsin (t/α) 

where x is its position in metre, t is time in s and α is a positive constant 

The particle comes to rest at time t = (in terms of α and π)

Asked By: SWATI KAPOOR
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Joshi sir comment

differentiating given x w. r. t. t 

dx/dt will be v 

for rest v = 0

Physics >> Mechanics part 1 >> Kinematics Medical Exam

a vector a which has magnitude 8 is added to a vector B which lies on x-axis. The sum of these two vectors, lies on y-axis and has a magnitude twice of the magnitude of B. The magnitude of vector B is

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Joshi sir comment

let a = 8cosAi+8sinAj and B = bi

sum of the two =  8cosAi+8sinAj+bi

it is given as 2bj  

so   8cosAi+8sinAj+bi = 2bj

so   (8cosA+b)i+8sinAj = 2bj

now on comparing coefficient of i and j we get

cosA = -b/8 and sinA = 2b/8

now use sin2A +cos2A = 1

 

Physics >> Mechanics part 1 >> Kinematics Medical Exam

the forces F1 and F2 are acting perpendicular to each other at a point and have resultant R. If force F2 is replaced by R2 - F12/F2 acting in the direction opposite to that of F2, the magnitude of resultant

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Joshi sir comment

F1+ F22 = R2  (1)

now for second condition

F1+ [(R2 - F12)/F2]2  = R12          R1 is new resultant

on putting the value of Rfrom first eq. we get

F1+ [ F22/F2]2 = R12

so R = R1

A sleeping cat and a sleeping dog are initially at a distance 'l' metres. As soon as the dog wakes up, the cat runs in the horizontal direction with constant velocity of u m/s. Immediately the dog chases the cat with constant speed of v m/s. Initially the dog was facing the north direction towards cat. The cat starts running in the east direction. Assuming the dog aims at the cat with constant speed v m/s, find the time after which the dog chases the cat.

Asked By: ADARSH
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Joshi sir comment

see video for this solution in you tube by the name iitbrain.com

video:  i e irodov 1.13 

Physics >> Mechanics part 1 >> Kinematics Medical Exam

Time of flight of a projectile thrown from ground is 8√5 s. It will rise up to a height of.........

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Joshi sir comment

T = 2usinθ/g 

by formula calculate usinθ

then put in H formula

Physics >> Mechanics part 1 >> Kinematics Medical Exam

A projectile is projected with velocity u= ai+bj from ground if acceleration due ti gravity is g, then the change in velocity of the projectile in second of projection is

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Joshi sir comment

for t sec of projection v = u + at so v = ai+bj-gtj

now calculate v-u then mod

Physics >> Mechanics part 1 >> Kinematics Medical Exam

A shell at rest at large height explodes into two parts which move horizontally with speeds v1 and v2. After what time their instantaneous velocities are perpendicular?

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Joshi sir comment

please see iitbrain video for the answer in you tube 

iitbrain.com in you tube 

video name i e irodov problem 1.11

Physics >> Mechanics part 1 >> Kinematics Medical Exam

A particle moving in uniform circular motion undergoes an angular displacement α (less then 360°) from point P and Q. Then angle between average velocity from P to Q and velocity at Q is........

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Joshi sir comment

as clear from the figure that angle is a/2

Physics >> Mechanics part 1 >> Kinematics Medical Exam

A particle moves in the plane xy with velocity v = ai + bxj, where i and j are the unit vectors of the x and y axes, and a and b are constants. At the initial moment of time the particle was located at the point x = y = 0. Find: 
(a) the equation of the particle's trajectory y(x); 
(b) the curvature radius of trajectory as a function of x. 

Asked By: SWATI KAPOOR
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Joshi sir comment

see the video in you tube by the name iitbrain.com 

video i e irodov 1.35

Physics >> Mechanics part 1 >> Kinematics Medical Exam

A ball is thrown up with some velocity after time t second it is at height h from the ground which is given by h = at -1/2 bt2. The dh/dt is zero at t equal to

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Joshi sir comment

dh/dt = a - bt 

so for dh/dt = 0 

t = a/b

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