302 - Mechanics part 1 Questions Answers
a particle initially at rest moves along x-axis. Its acceleration is given as a = (2t+2) m/s2. If it starts from origin , find distance covered in 2 second.
dv/dt = 2t+2
so dv = 2tdt+2dt
on integrating the given eq.
v = t2 + 2t + C
at start v = 0 at t = 0
so on puting these initial values we get C = 0
so v = t2 + 2t
so dx/dt = t2 + 2t
so dx = t2dt + 2tdt
on integrating x = t3/3 + t2+ C
on putting initial values x=0 at t=0
we get C = 0
so x = t3/3 + t2
now put t = 2
Angle between a and b is θ. What is the value of a.(bxa) ?
If two vectors are some in scalar triple product then it will be 0 always
the velocity v of a particle moving in the positive direction of x - axis as v = 5√x, assuming that at t=0, particle was at x=0. What is the acceleration of the particle?
v = 5√x
differetiate this we get
dv/dt = 5*1/(2√x)*dx/dt
so a = 5*1/(2√x)*v
so a = 5*1/(2√x)*5√x
so a = 12.5
a man wants to cross a river of width 500m in minimum distance. The rowing a speed of man relative to water is 3 km/h and river flows with speed 2km/h. The time taken by man to cross the river is
for minimum drift he should swim in a direction such that 3sinA = 2
so sinA = 2/3
now calculate cosA
then time = 500/3cosA
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)
differentiating given x w. r. t. t
dx/dt will be v
for rest v = 0
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
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
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
F12 + F22 = R2 (1)
now for second condition
F12 + [(R2 - F12)/F2]2 = R12 R1 is new resultant
on putting the value of R2 from first eq. we get
F12 + [ 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.
see video for this solution in you tube by the name iitbrain.com
video: i e irodov 1.13
Sir, can third law of Newton be derived from second law??
actually as we all know that second law is real law so first & second law can be derived, but on the other side it is written that they are THE LAWS ARE INDEPENDENT THUS CAN'T BE DERIVED FROM ANY OTHER LAW (acc to HC Verma)
sir sometime ans is yes while sometime no , in different books different ans , i consulted this query to many other teachers also but some favours while some teachers not , so what you suggest?
is this depends upon questn setting person
THE LAWS ARE INDEPENDENT
distance covered in n sec. = un+an2/2 (1)
distance covered in n-1 sec. = u(n-1)+a(n-1)2/2 (2)
so sn = (1) - (2)
put equation (1) and (2) and get the answer