If a_{r} and a_{t} represent radial and tangential accelerations, the motion of a particle will be uniformly circular if :
a_{r} = 0 and a_{t} = 0
a_{r} = 0 and a_{t} ≠ 0
a_{r} ≠ 0 and a_{t} = 0
a_{r} ≠ 0 and a_{t} ≠ 0
C.
a_{r} ≠ 0 and a_{t} = 0
If a_{r} = 0 and a_{t} = 0, then motion is uniform translatory.
If a_{r} = 0 but a_{t} ≠ 0, then motion is accelerated translatory.
If a_{r} ≠ 0 but a_{t} = 0, then motion is uniform circular.
If a_{r} ≠ 0 and a_{t} ≠ 0, then motion is a non-uniform circular.
A particle starting from the origin (0,0) moves in a straight line in the (x,y) plane. Its coordinates at a later time are ( ). The path of the particle makes with the x -axis an angle of:
30^{o}
45^{o} C
60^{o} C
0^{0}
C.
60^{o} C
The slope of the path of the particle gives the measure of angle required.
Draw the situation as shown. OA represents the path of the particle starting from origin O (0,0). Draw a perpendicular from point A to X- axis. Let path pf the particle makes and angle θ with the x -axis, then
tan θ = slope of line OA
For angles of projection of a projectile at angles (45° - θ) and (45° + θ), the horizontal ranges described by the projectile are in the ratio of
1:1
2:3
1:2
2:1
A.
1:1
For complementary angles of projection, their horizontal ranges will be same.
We know that, horizontal ranges for complementary angles of projection will be same.
The projectiles are projected at angles and which are complementary to each other i.e., two angles add up to give . Hence, horizontal ranges will be equal. Thus, the required ratio is 1:1.
A car runs at aconstant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap respectively is
0, 0
10 m/s, 10m/s
10 m/s, 0
B.
In completing a circular lap, car's displacement is zero.
Average velocity is defined as the ratio of displacement to time taken while the average displacement to time taken while the average speed of a particle in a given interval of time is defined as the ratio of distance travelled to the time taken.
On a circular path in completing one turn, the distance travelled is while displacement is zero.
Hence, average velocity =
A car moves from X to Y with a uniform speed v_{u} and returns to Y with a uniform speed v_{d}. The average speed for this round trip is:
A.
The average speed of a body in a given time interval is defined as the ratio of distance travelled to the time taken.
Let t_{1} and t_{2} be times taken by the car to go from X and Y and then Y to X respectively.
Then,
60^{o}
45^{o}
30^{o}
90^{o}
A.
60^{o}
The vectors are such that a:
The angle between the two vectors is
A.
As we have given
where is the angle between
Squaring both sides, we have
Hence, angle between
The position x of a particle with respect to time t along x- axis is given by x = 9t^{2} -t^{3} 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
24 m
B.
54 m
At the instant when speed is maximum, its acceleration is zero.
Given, the position x of particle with respect to time t along x- axis
x = 9t^{2}-t^{3} ... (i)
differentiating Eq. (i) with respect to time, we get speed, ie,
Again differentiating Eq. (ii) with respect to time, we get acceleration, ie,
Now, when speed of particle is maximum, its acceleration is zero, ie,
a= 0
18-6t = 0 or t = 3s
Putting in eq (i) We, obtain the position of a particle at that time.
x = 9 (3)^{2} - (3)^{3} = 9 (9) -27
= 81-27 = 54 m
A particle moving along x- axis has acceleration f, at time t, given f = f_{o} (1-t/T), where f_{o} and T are constants. The particle at t =0 and the instant when f = 0, the particle's velocity (v_{x}) is:
f_{o}T
f_{o}T^{2}/2
f_{o}T^{2}
f_{o}T/2
D.
f_{o}T/2
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
19/3
C.
10/3
Distance travelled by the particle in nth second
Where u is initial speed and a is an acceleration of the particle.
Here, n = 3, u =0, a = 4/3 m/s^{2}