114 - Kinematics Questions Answers
A particle moving with constant acceleration covers a distance of 30m in the 3rd second. it covers a distance of 50m in the 5th second. What is the acceleration of the particle?
use formula for the distance covered in a particular second
st = u+at-(a/2)
A particle is moving with a velocity of 10m/s due east. In One second its velocity changes to 10m/s due west. If the particle is uniformly accelerated, the change in velocity will be directed as
let east is i
then velocity change = -10i-10i = -20i
A ball is dropped from a height h reaches the ground in time T. What is its height at time T/2
h = 1/2 gT2
in T/2 time h' = 1/2 gT2/4 = 1/8 gT2
so height from ground = h-h'
solve?
see you tube video of iitbrain by the name I E Irodov problem 1.3
Two Particles move in a uniform gravitational field with an acceleration g. At The Initial moment the particles were located at one point and moved with velocities v1= 3.0ms-1 and v2 = 4.0ms-1 horizontally in opp. Direction.Find The distance between the particles at the moment when their velocity vector become mutually Perpendicular?
see you tube video of iitbrain by the name I E Irodov problem 1.11
A Train Accelerates from rest at a constant rate α for sometime and then it retards to rest at a constant rate β. If te total distance covered by the particle is x, then what is the maximum velocity of the train ?
for first part of motion
v2 = 2αx1 (1)
similarly for second part of motion
v2= 2βx2 (2)
now by given conditions
x = x1 + x2
so x = v2/2α + v2/2β
so x = v2/2[1/α + 1/β]
now solve
A taxi leaves the station X for Station Y every 10 min. Simulataneously, a taxi also leaves the station Y for station X Every 10min. The Taxis move at the same constant speed and go from X to Y or Viceversa in 2Hours. How Many taxis coming from the other side meet each taxi enroute from Y To X?
the answer depends on the condition that where taxi meets with the first taxi coming from opposite side.
If we consider that at the starting point taxi meets with the first taxi coming from other side, then after every 5 min it will meet with a taxi from opposite side. This will be due to the double relative speed of two taxi moving opposite. Total time length of the path = 2 hrs = 120 min
so no. of taxi = 120/5 + 1 = 25
if first taxi meet at the point a little ahead the starting point then no of taxi = 120/5 = 24
similarly other cases
A train accelerates from rest at a constant rate α for distance x1 and time t1.After that it retards to rest at constant rate β for distamce x2 and time t2.Which of the following relation is correct?
| A. x1/x2=α/β=t2/t1 | B. x1/x2=β/α=t2/t1 |
| C. x1/x2=α/β=t1/t2 | D. x1/x2=β/α=t1/t2 |
for first part x1 = αt12/2 and v = αt1
similarly for second part x2 = βt22/2 and v = βt2
on comparing v we get αt1 = βt2
so β/α = t1/t2 (1)
and x1/x2 = αt12/βt22 = β/α (2) by using (1)
let tangential acc is a
then velocity after covering half circle
v2 = 2a(πr)
so centripetal acc = v2/r = 2aπr/r = 2aπ
so angle = tan-1[2aπ/a] = tan-12π
here notice that velocity will be in the direction of tangential acc.
For what angle should a body be projected with velocity 24m/s just to pass over the obstacle 16 m high at horizontal distance of 32 m? Take g=10 m/s^2
use the eq.
y = xtanθ-(gx2/2u2cos2θ)