Motion In One Dimension Test

Result

Motion In One Dimension Test - 6

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Weekly Quiz Competition

Question 1
4 / -1

An insect moving along a straight line, travels in even seconds distance equal to the magnitude of time elapse. Assuming acceleration to be constant, and the insect standard at t=0. Find the magnitude of initial velocity of insect :

A
3/2 unit

B
5/4 unit

C
2 unit

D
1/2 unit

Solution

Lets consider a insect moving with initial velocity u and a distance covered from ′t−1′ to ′t′ is t
Question 2
4 / -1

Two bikes A and B start from a point. A moves with uniform speed 40m/s and B starts from rest with uniform acceleration 2m/s² . If B starts at t=10 and A starts from the same point at t=10s, then the time during the journey in which A was ahead of B is

A
20s

B
8s

C
10s

D
A is never ahead of B

Solution
Question 3
4 / -1

Three forces start acting simultaneously on a particle moving with velocity . These forces are represented in magnitude and direction by the three sides of a triangle ABC(as shown). The particle will now move with velocity

A
less than

B
greater than

C
in the direction of the largest force BC

D
, remaining unchanged.

Solution

According to triangle law of vector addition if three vectors addition if three vectors are represented by three sides of a triangle taken in same order, then their resultant is zero. Therefore resultant of the forces acting on the particle is zero, so the particles velocity remains unchanged.
Question 4
4 / -1

A body starts from rest with an uniform acceleration. If its velocity after n seconds is v, then its displacement in the last 2s is

A

B

C

D

Solution
Question 5
4 / -1

A particle is moving in a straight line. and passes through a point o with a velocity of 6 ms^-1.The particle moves with a constant retardation of 2ms^-2 for 4 s and there after moves with constant velocity. How long after leaving O does the particle return to O?

A
3s

B
8s

C
Never

D
4s

Solution

Let the particle moves toward right with velocity 6 m/s. Due to retardation, after time t₁, its velocity becomes zero.

But retardation works on it for 4 sec. It means after reaching point A, direction of motion gets reversed and acceleration works on the particle for next one second.
Question 6
4 / -1

A projectile is fired vertically upwards with an initial velocity u. After an interval of T seconds, a second projectile is fired vertically upwards, also with initial velocity u.

A

B

C

D
They never meet

Solution
Question 7
4 / -1

A ball is dropped vertically from a height d above the ground it hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistances, its velocity v varies with the height h above the ground as

A

B

C

D

Solution

Using the equation v²−u²= 2as

We get v² = 2as Since u=0

Clearly the graph will be parabolic.

The velocity will be negative while the ball falls while it will be positive when it bounces back i.e. they will have opposite direction, also the maximum height will be 2/d
Question 8
4 / -1

Two trains, which are moving along different tracks in opposite directions, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given in figure show their velocities as function of time as the trains slow down. The separation between the trains when both have stopped is

A
120 m

B
280 m

C
60 m

D
20 m

Solution
Question 9
4 / -1

A drunkard is walking along a straight road. He takes five steps forward and three steps backward and so on. Each step is 1 m long and takes 1 s. There is a pit on the road 11 m away from the starting point. The drunkard will fall into the pit after

A
29 s

B
21 s

C
37 s

D
31 s

Solution

Since the last five steps covering 5 m land the drunkard fell into the pit, the displacement prior to this is (11 - 5) m=6 m. Time taken for first eight steps (displacement in first eight steps = 5 - 3 = 2 m) = 8 s. Then time taken to cover first 6 m of journey = 6/2 × 8 = 24s

Time taken to cover last 5 m = 5 s

Total time = 24 + 5 = 29 s
Question 10
4 / -1

A ball is thrown from the top of a tower in vertically upward direction. The velocity at a point h meter below the point of projection is twice of the velocity at a point h meter above the point of projection. Find the maximum height reached by the ball above the top of tower.

A
2h

B
3h

C
(5/3)h

D
(4/3)h

Solution