A body is at rest at x = 0 and t = 0, it starts moving in the positive X-direction with a constant acceleration. At the same instant, another body passes through x = O moving in the positive X-direction with a constant speed. The position of the first body is given by x1 (t) after time t and that of second body by x2 (t) after the same time interval. Which of the following graphs correctly describes (x1 − x2) as a function of time t ?
The given graph shows the variation of velocity with displacement. Which one of the graphs given below correctly represents the variation of acceleration with displacement.
A.
Given line has positive intercept but negative slope. So, its equation can be written as
v = − mx + V0
By differentiating w.r.t. to time, we get
Now, substituting the value of v from Eq. (i), we get
i.e., the graph between a and x should have positive slope but negative intercept on a-axis. So, graph (a) is correct.
In figure shown below, a particle is travelling counter clockwise in circle of radius 10 m. The acceleration vector indicated at a specific time. Find the value of 'v' at this time.
10 m/s
15 m/s
20 m/s
7 m/s
A ball which is at rest is dropped from a height h metre. As it bounces off the floor its speed is 80% of what it was just before touching the ground. The ball will then rise to nearly a height
0.94 h
0.80 h
0.75 h
0.64 h
Two bodies of different masses are dropped from heights of 16 m and 25 m respectively. The ratio ofthe time taken by them to reach the ground is
A man of height h walks in a straight path towards a lamp post of height H with uniform velocity u. Then the velocity of the edge of the shadow on the ground will be
A train of 150 m length is going towards north direction at a speed of 10 ms-1 . A parrot flies at a ,speed of 5ms-1 towards south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to
12 s
8 s
15 s
10 s
The area under acceleration-time graph gives
distance travelled
change in acceleration
force acting
change in velocity
The distance travelled by an object along a straight line in time t is given by s= 3 − 4t + 5t2, the initial velocity of the object is
3 unit
− 3 unit
4 unit
− 4 unit
A rocket of mass 100 kg burns 0.1 kg of fuel per sec. If velocity of exhaust gas is 1 km/sec, then it lifts with an acceleration of
1000 ms-2
100 ms-2
10 ms-2
1 ms-2