575 - Physics Questions Answers
The displacement x of a particle varies with time t as x2= 1+t2 . What is its acceleration?
on diffrentiation we will get
2xdx/dt = 2t
so xv = t (1)
on differentiating again
xdv/dt + vdx/dt = 1
or xa + v2 = 1 (2)
now put v from (1) to (2) we will get answer
a particle moving with constant acceleration on a straight line ultimately comes to rest. What is the angle between its initial velocity and acceleration
180 degree because it is retarded particle
a particle moves along a straight line. Its position (x) at any instant is given by x= 32t - 8t3/3, where x is in metre and t in second. Find the acceleration of the particle at the instant particle comes to rest
differentiate the given function and find v, again differentiate find a
compare v to 0 for rest, get t and put in a
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