A particle covers half of its total distance with speed v1 and the rest half distance with speed v2. Its average speed during the complete journey is
v1v2v1+v2
2v1v2v1+v2
2v12v22v12+v22
v1+v22
A ball is dropped from a high rise platform at t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18 s. What is the value of v? (take g=10 ms-2)
75 ms-1
40 ms-1
A particle moves a distance x in time t according to equation x=t+5-1. The acceleration of the particle is proportional to,
velocity3/2
A bus is moving with a speed of 10 ms -1 on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what minimum speed should the scooterist chase the bus?
20 ms-1
25 ms-1
10 ms-1
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
The distance travelled by a particle starting from rest and moving with an acceleration 43ms-2, in the third second is
6m
4m
103m
193m
A particle shows a distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point:
B
C
D
A
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 of 135 m in t seconds. The value of t is:
10
1.8
12
9
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t1. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t2. The time taken by her to walk upon the moving escalator will be:
t1t2t2-t1
t1t2t2+t1
t1-t2
t1+t22
The coordinate of an object is given as a function of time by x=7t-3t2, where x is in meters and t is in seconds. Its average velocity over the interval t=0 to t=4 is will be
5 m/s
-5 m/s
11 m/s
-11 m/s
A particle moves along a straight line and its position as a function of time is given by x=t3-3t2+3t+3, then the particle:
stops at t=1 sec and reverses its direction of motion.
stops at t=1 sec and continues further without a change of direction.
stops at t=2 sec and reverses its direction of motion.
stops at t=2 sec and continues further without a change of direction.
A particle moves in a straight line, according to the law x=4at+a sinta , where x is its position in meters, t is in sec & a is some constant, then the velocity is zero at :
x=4a2π meters
t=π sec
t=0 sec
none
A point moves in a straight line so that its displacement is x m at time t sec, given by x2=t2+1. Its acceleration in m/s2 at time 1 sec is:
1x
1x-1x3
1x3
-t2x3
If the velocity of a particle is v =At + Bt2, where A and B are constants, then the distance travelled by it between 1s and 2s is:
3A+7B
32A+73B
A2+B3
32A+4B
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) =βx-2n where, β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by-
A particle is moving such that its position coordinates (x, y) are (2m, 3m) at time t = 0, (6m,7m) at time t = 2s and (13m, 14m) at time t = 5 s, Average velocity vector v→avgfrom t = 0 to t = 5 s is :
15(13i^+14j→)
73(i^+j→)
2(i^+j→)
115(i^+j→)
A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h1, h2 and h3 is :
h1 = h23 = h35
20 m/s
40 m/s
10 m/s
A particle moves a distance x in time t according to equation X = (t + 5)-1 The acceleration of particle is proportional to
(velocity)3/2
(distance)2
(distance)-2
(velocity)2/3
A particle starts its motion from rest under the action of a constant force. If the distance covered in the first 10 sec is S1 and that covered in the first 20 sec is S2, then
S2 = 2S1
S2 = 3S1
S2 = 4S1
S2 = S1
A particle shows the distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point:
The position x of a particle with respect to time t along the x-axis is given by x = 9t2 – t3 where x is in metre 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
A car moves from X to Y with a uniform speed vu and returns to X with a uniform speed vd. The average speed for this round trip is :
2vdvuvd+vu
vuvd
vdvuvd+vu
vu+vd2
A particle moving along the x-axis has acceleration f, at time t, given by f=f01-tT, where f0 and T are constants. The particle at t=0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle’s velocity (vx) is:
f0T
12f0T2
f0T2
12f0T
A particle moves along a straight line OX. At a time t (in seconds), the displacement x (in metres) of the particle from O is given by x= 40 + 12t – t3. How long would the particle travel before coming to rest?
40 m
56 m
16 m
Two bodies, A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is:
5/4
12/5
5/12
4/5
For acceleration of a particle varies with time as shown in the figure. Calculate the displacement of the particle in the time interval from t=2 s to t=4 s.
203m
20 m
10 m
Referring to a-s diagram as shown in the figure. Find the velocity of the particle when the particle just covers 20 m (v0=50m s-1).
350m s-1
300m s-1
100m s-1
10m s-1
A point starts moving in a straight line with a certain acceleration. At a time 't' after beginning of motion the acceleration suddenly becomes retardation of the same value. The time in which the point returns to the initial point is-
2t
2+2t
t2
Cannot be predicted unless acceleration is given
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