114 - Kinematics Questions Answers
a body is p rojected vertically upwards with speed 20m/s .find the distance travelled by it during the last second of its upward journey(takeg=10m/s2
last sec. of upward journey means first sec. of free fall from the top so s=0(1) + 1/2(10)(1)2 = 5 meter, here g is taken 10m/s2
The velocity of a body moving along x- axis is given by v =√( 36+2x) where v is in m/s and x is in metre. Find the distance travelled by it in 2 nd second.
The answer is 7.5 m
dx/dt = √( 36+2x)
so dx/√( 36+2x) = dt
now integrate
limits of x are x1 to x2 and that of time are 1 to 2
एक नाव की शांत जल में चाल 5 km /h है ,तथा 1 km चोडी नदी को न्यूनतम पथ के अनुदिश 15 मिनट में पार करती है तो नदी के जल प्रवाह का वेग km/h होगा..?
let boat is moving in a direction making an angle θ with the perpendicular to the flow, then for minimum drift
5sinθ = r
and 5cosθ = 1/.25 = 4
so cosθ = 4/5
and sinθ = 3/5
so r = 3 km/hr
You can see video related to this question in you tube (iit brain) as the name irodov problems
Velocity of raindrops in still air is 4 m/s vertically downward. If wind starts blowing at the rate 3 m/s horizontally towards north, then find the direction along which a man standing on the ground would keep his umbrella ti protect himself from rain?
sir, i dont know the figure of this ques, so plz tell me this ques diagramatically.
angel of umbrella with vertical = tan-1[3/4]
d benches of a gallery in a cricket stadium r 1m wide & 1m high . a boy strikes d ball at a level of 1m above d ground & hits a sixer. d ball starts at 35m/s at an < of 530 with d horizontal . d benches r perpendicular to plane of motion & d first bench is 110m from d boy. ON WHICH BENCH WILL THE BALL HIT??
ANS 6TH
we know that y = xtanθ - gx2/2u2cos2θ
according to the given conditions
x = 110+n
y = n here n is the number of bench in which ball strikes
θ, u and g are given
A particle starts from rest and moves with constant acceleration. The magnitude of its displacement is
1) more than its distance traversed
2) less '' '' '' ''
3) equal to '' '' ''
4) zero
ans given is (3) but I want to know why it vyl not be less than dist. travelled & why (4) optn is wrong as in case of uniform circular motion displacement is 0 but have const. acc.??
I think the particle is moving in a straight line
A car moves towards north at a speed of 54 km/h for 1 h. Then it moves eastward with same speed for same duration. The average speed and velocity of car for complete journey is
avg speed = total distance / total time = 54+54/1+1 = 54
avg velocity = total displacement / total time = 54sqrt(2) / 2 = 27sqrt(2)
The acceleration (a) of a body moving along x-axis is given by a=4x3 ,where a is in m/s2 and x is in metre.if at x=0 the velocity of body is 2m/s,then find its velocity at x=4m?
a = 4x3
so dv/dt = 4x3
or dv/dx * dx/dt = 4x3
or dv/dx * v = 4x3
or vdv = 4x3dx
or ∫vdv = 4∫x3dx
or v2/2 = 4x4/4 + C
at x = 0, v = 2
so 4/2 = 0+C
so C = 2
on putting the value of C
we get v2/2 = x4+ C
now put x = 4 and get v
The position x of a particle moving along x-axis at time t is given by equation t=√x +4 , where x is in meter and t is in seconds. Find the position of particle when its velocity is zero.
given that x = (t-4)2
so dx/dt = 2(t-4)
or v = 2t-8
now for v=0, t=4
put t=4 and get x
A body projected upwards is at same height form ground at t=3s and t=7s . Find the maximum height attained by it. ( take g= 10 m/s^2 ).
A body projected upwards is at same height form ground at t=3s and t=7s. It will gain max. height after 5 second
so maximum height = gt2/2 = 10*5*5/2 = 125
another direct method for finding total distance = 5+15+25+35+45 = 125
these are the distances in 1st, 2nd, 3rd and etc sec.