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Test: Motion in a Straight Line (May 23) - NEET MCQ


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12 Questions MCQ Test Daily Test for NEET Preparation - Test: Motion in a Straight Line (May 23)

Test: Motion in a Straight Line (May 23) for NEET 2024 is part of Daily Test for NEET Preparation preparation. The Test: Motion in a Straight Line (May 23) questions and answers have been prepared according to the NEET exam syllabus.The Test: Motion in a Straight Line (May 23) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Motion in a Straight Line (May 23) below.
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Test: Motion in a Straight Line (May 23) - Question 1

The displacement x of a particle depend on time t as x = at2 - bt3

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 1

Step 1: Velocity

Velocity v is the first derivative of displacement x with respect to time t:

Step 2: Time when the particle comes to rest

A particle comes to rest when its velocity v=0. Setting v to zero:

2at − 3bt= 0

 t (2a−3bt) = 0

This gives two solutions:

Thus, the particle comes to rest after time t = 2a / 3b.

 

Step 3: Initial Velocity and Acceleration

  • Initial velocity v0​ is obtained by substituting t=0 in the velocity equation:

So, the initial velocity is zero.

  • Acceleration anet​ is the derivative of velocity:

At t=0, the acceleration is:

Thus, the initial acceleration is not zero.

Step 4: Net Force

The net force acts based on acceleration. Given the acceleration:

Test: Motion in a Straight Line (May 23) - Question 2

 A particle has intial velocity 10 m/s. It moves due to constant retarding force along the line of velocity which produces a retardation of 5 m/s2. Then -

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 2

 

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Test: Motion in a Straight Line (May 23) - Question 3

A particle initially at rest is subjected to two forces. One is constant, the other is a retarding force proportion at to the particle velocity. In the subsequent motion of the particle

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 3

This represents a case that is similar to that of a raindrop attaining terminal velocity.
Fnet = (k1 – k2v)
The acceleration wil become 0 and the velocity will be constant eventually.
 

Test: Motion in a Straight Line (May 23) - Question 4

Let v and a denote the velocity and acceleration respectively of a body

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 4

A can be zero when velocity and vice-versa you can get that when you see a pendulum at mean a = 0 but v = max and at extreme a = max when v = 0. Also a “can” be zero when v is zero.

Test: Motion in a Straight Line (May 23) - Question 5

The figure shows the velocity (v) of a particle plotted against time (t)

                  

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 5
  1. As it crossed the x axis so the direction changed
  2. The slope of v-t graph gives acceleration so it is constant
  3. The net areas below and above x-axis are equal so the displacement is zero
  4. Speed is a scalar quantity hence the initial and final speeds are same
Test: Motion in a Straight Line (May 23) - Question 6

A particle moves with constant speed v along a regular hexagon ABCDEF in the same order. Then the magnitude of the average velocity for its motion from A to -

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 6

 

Test: Motion in a Straight Line (May 23) - Question 7

 A projectile of mass 1 kg is projected with a velocity of  m/s such that it strikes on the same level as the point of projection at a distance of  m. Which of the following options are incorrect.

Test: Motion in a Straight Line (May 23) - Question 8

Choose the correct alternative.

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 8

Let A be the angle of projection and u be the velocity of projection.

 The displacement in y-direction is zero.
0 = (u sin q)t - gt²/2
t = 2u sin q/g
R = (2u²sin q cos q)/g
2u²sin q cos q = gR
Multiply by 2 tan q
4u²sin²q = 2gR tan q
(2u sin q)² = 2gR tan q
g²t²= 2gR tan q
gt²= 2R tan q

Test: Motion in a Straight Line (May 23) - Question 9

The a-t graph of the particle is correctly shown by

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 9

The acceleration of the particle moving in straight line is given as, a=d2x​/dt2
Geometrically it is the concavity of the x−t graph. If the graph is concave upward then the double derivative is positive. If the graph is concave downwards then the double derivative is negative.Consider the provided x−t graph,
Since the x−t graph is a parabola which of second degree the a−t graph must be a constant line which is of zero degree. Since the parabola is concave downwards the acceleration must be negative.
The only graph which satisfies the above conditions is graph D.


Hence, option D is correct.

Test: Motion in a Straight Line (May 23) - Question 10

The figure shows a velocity-time graph of a particle moving along a straight line

         

Choose the incorrect statement. The particle comes to rest at

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 10

The particle comes to rest (v = 0) at two instants. It comes to rest first time at somewhere between 4th second and 6th second. At t = 8 second it again comes to rest.
We will find when the particle comes to rest for the first time. Between 4th and 6th second (time interval of 2 seconds) the particle changes it's velocity from 10 m/s to -20 m/s. So, the acceleration on the particle during this part of motion is,
a = (-20 - 10)/2 = -15 seconds
The particle comes to rest when v = 0
v - u = at
0 - 10 = (-15)t
⇒ t = 1.5 seconds
So the particle come to rest at (4 + 1.5)th = 5.5 th second and 8th second

Test: Motion in a Straight Line (May 23) - Question 11

A train covers 60 miles between 2 p.m. and 4 p.m. How fast was it going at 3 p.m.?

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 11

The speed is traveled distance (60 miles) divided by traveled time (4pm – 2pm = 2hours):
60 miles/ 2 hours = 30 mph

Test: Motion in a Straight Line (May 23) - Question 12

The figure shows a velocity-time graph of a particle moving along a straight line

         

The maximum of displacement of the particle is

Detailed Solution for Test: Motion in a Straight Line (May 23) - Question 12

The area under the v-t graph gives displacement. The maximum displacement is when the particle velocity becomes zero and it changes its direction. Since we don't know when the particles crosses x-axis (v = 0) we will first find displacement till 4 seconds
S = (1/2)(10)(2) + (10)(2) = 30 m
The particle velocity changes from 10 m/s to - 20 m/s in next 2 seconds. So it's acceleration during this part of motion is,
a = (-20 - 10)/2 = - 15 m/s²
Distance travelled till velocity becomes zero can be calculated by using the formula v² - u² = 2as
0 - (10)² = 2(-15)s
=> s = 100/30 = 3.33 m
Therefore the maximum displacement is,
d = 30 + 3.3 = 33.3 m
 

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