CBSE
A particle shows distance -time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point
B
C
D
D
Two bodies of mass 1 kg have position vectors, respectively.The centre of mass of this system has a position vector
A particle moves in the x-y plane according to rule x = a sin ωt and y = a cos ωt. The particle follows
an elliptical path
a circular path
a parabolic path
a parabolic path
A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms-1 to 20 ms-1 while passing through a distance 135 m in t second. The value of t is
10
1.8
12
12
The speed of the projectile at its maximum height is half of its initial speed. The angle of projection is
60o
15o
30o
30o
The position x of a particle with respect to time t along x- axis is given by x = 9t2 -t3 where x is in meter and t in second. What will be the position of this particle when it achieves maximum speed along the +x direction?
32 m
54 m
81 m
81 m
The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms-2 in the third second is
6 m
4 m
10/3
10/3
A particle of mass is projected with the velocity v making an angle of 45o with the horizontal.
When the particle lands on the level ground the magnitude of the change in its momentum will be
2 mv
If is the force acting on a particle having position vector
be the the torque of this force about the origin, then
A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 s is s1 and that covered in the first 20 s is s2, then
s2 = 2s1
s2 = 3s1
s2 = 4 s1
s2 = 4 s1