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Test: Motion in a Plane and with Constant Acceleration (May 15) - NEET MCQ


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10 Questions MCQ Test - Test: Motion in a Plane and with Constant Acceleration (May 15)

Test: Motion in a Plane and with Constant Acceleration (May 15) for NEET 2024 is part of NEET preparation. The Test: Motion in a Plane and with Constant Acceleration (May 15) questions and answers have been prepared according to the NEET exam syllabus.The Test: Motion in a Plane and with Constant Acceleration (May 15) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Motion in a Plane and with Constant Acceleration (May 15) below.
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Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 1

A bird flies from (-3m, 4m, -3m) to (7m, -2m, -3m) in the xyz- coordinates. The bird's displacement vector is given by

Detailed Solution for Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 1

Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 2

If x = 5t + 3t2 andy = 4t are the x and y co-ordinates of a particle at any time t second where x and y are in metre, then the acceleration of the particle

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Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 3

The (x, y, z) coordinates of two points A and B are given respectively as (0, 4, -2) and (-2, 8, -4). The displacement vector from A to B is:

Detailed Solution for Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 3


 Displacement vector from A to B is given by

Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 4

The position of a particle is given by where t is in seconds and the coefficients have the proper units forto be in metres. The direction of velocity of the particle at t = 1 s is

Detailed Solution for Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 4


Let θ be angle which the direction ofmakes with the

Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 5

The acceleration of the particle at t = 1 is

Detailed Solution for Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 5

From the previous question, 

Acceleration of the particle remains constant at all times.

Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 6

For any arbitrary motion in space, which of the following relations is true?

Detailed Solution for Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 6

The relation (b) is true, others are false because relations (a), (c) and (d) hold only for uniformly accelerated motion.

Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 7

Four girls skating on circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in figure. For which girl displacement is equal to the actual length of path?

Detailed Solution for Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 7

Displacement of each girl = 
Magnitude of displacement for each girl
For girl C, displacement = actual length of path.

Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 8

A particle starts moving from point (2, 10, 1). Displacement for the particle is The final coordinates of the particle is

Detailed Solution for Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 8

Initial position vector of the particle,
Let final position vector of the particle be

Hence, the final coordinates of the particle are ((10, 8, 2)

Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 9

On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, the displacement of the motorist at the third turn is

Detailed Solution for Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 9

The path followed by the motorist is a regular hexagon ABCDEF of side 500 m, as shown in the given figure. Let the motorist starts from A and takes the third turn at D.
image
Therefore, the magnitude of this displacement is 
​AD = AG + GD = 500 m + 500 m = 1000 m 

Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 10

A person moves 30 m north, then 30 m east, then 30√2​ m  south-west. His displacement from the original position is:

Detailed Solution for Test: Motion in a Plane and with Constant Acceleration (May 15) - Question 10

Resolving displacement 30√2 m south-west into two rectangular components, we get
Displacement in south

∴ Effective displacement due to 30 m north and 30 m south is zero.
Also effective displacement due to 30 m east and 30 m west in zero.

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