114 - Kinematics Questions Answers
A man moves in an open field such that after moving 10m on a straight line, he makes a
sharp turn of 60° to his left. Find the displacement after 8 such turns
draw a hexagon
just after 8 such turn displacement = 10√3
a particle is moving eastwards with a speed of 6m/s. After 6s, the particle is found to be moving with same speed in a direction 60° north of east. The magnitude of average acceleration in this interval of time is ..............
initial velocity = 6i
final velocity = 6cos60i+6sin60j
now solve
a body is dropped from a certain height h (h is very large) and second body is thrown downward with velocity of 5 m/s simultaneously. What will be difference in heights of the two bodies after 3s.
g is acting on both bodies so difference = 5*3 = 15 m
a stone thrown upward with a speed u from the top of a tower reaches the ground with a velocity 4u. The height of the tower is .........
(4u)2 = (-u)2+ 2gh
solve
The velocity of a body depends on time according to the equation v = (t2/10) + 20 . The body is undergoing -
1. Uniform acceleration
2. Uniform retardation
3. Non-uniform acceleration
dv/dt = 2t/10
so a depends on t
option (3)
When a particle is thrown vertically upwards, its velocity at one third of its maximum height is 10√2 m/s. The maximum height attained by it is......
remaining height = 2h/3
so 02 = (10√2)2 - 2g(2h/3)
solve
A body falling from a vertical height of 10m pierces through a distance of 1m in sand. It faces an average
retardation in sand equal to?
[Answer is in terms of 'g' i.e acceleration due to gravity ]
velocity at the time of impact
v2 = 02 + 2g10
now in sand this velocity becomes 0 in 1 m
so 02 = 20g - 2a(1)
so a = 10g
a body thrown vertically up with initial velocity 52 m/s from the ground passes twice a point at h height above at an interval of 10s. The height h is...........
(g=10m/s2)
from that point to topmost point, time taken = 10/2 = 5 sec
so velocity at that point = 50 m/s
now use
522 = 502- 2gh
solve
A bus starts moving with an acceleration of 2m/s2. A cyclist 96m behind the bus starts simultaneously towards the bus at 20 m/s . After what time cyclist will be able to overtake the bus?
let after time t
then distance covered by cyclist in this time = 20t
and by bus = 2t2 / 2 = t2
so according to the given condition
20t = 96+t2
so t = 8,12
after 8 sec cyclist overtakes bus and after 12 sec bus overtakes cyclist
a boat is moved uniformily by pulling with force p=240n,q=200n .what must be the inclination of the resultent force with p and q to have resultent r=400.
In this problem r is the resultant of p and q, let the angle between r and p is α and r and q is β then
p = r sin β/sin(α+β) (1) and q = r sinα/sin(α+β) (2)
so p/q = sinβ/sinα
and according to given condition p/q = 6/5 so sinβ = 6k and sinα = 5k
now obtain cos function and solve it.