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

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 − 3bt² = 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 a_net​ 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:

In this situation, two forces act on the particle:

1. A constant force, which will cause a constant acceleration and increase the particle's velocity over time.
2. A retarding force that is proportional to the particle's velocity, i.e., the retarding force is of the form -k * v, where k is a constant and v is the velocity of the particle.

Initially, when the particle is at rest, the retarding force is zero because the velocity is zero. As the particle starts moving, the constant force accelerates it, and the retarding force, which is proportional to velocity, starts acting in the opposite direction.

Over time:

• The constant force will continue to accelerate the particle, increasing its velocity.
• As the velocity increases, the retarding force also increases, opposing the motion, causing a reduction in acceleration.
• Eventually, the particle will reach a point where the net force is zero, meaning the acceleration decreases to zero, and the velocity stabilizes at a constant value.

Thus, the acceleration will decrease from its initial value (due to the constant force) to zero (due to the increasing retarding force), and the velocity will increase from zero to a constant value.

Therefore, the correct answer is:

Answer: B
The acceleration will decrease from its initial value to zero and velocity will increase from zero to a constant value.

Let's analyze each option:

A: A can be non-zero when v = 0 and A may be zero when v ≠ 0.

This is true.

• A can be non-zero when v = 0: A body can have a non-zero acceleration even when its velocity is zero, such as in the case of an object at rest that is about to start moving under the influence of a force (like a body dropped from rest under gravity, which initially has zero velocity but has a non-zero acceleration).
• A may be zero when v ≠ 0: A body can have a constant velocity (i.e., no acceleration) even if it's moving, such as an object moving at a constant speed in a straight line.

Hence, A is correct.

B: A must be zero when v = 0.
This is false.
The acceleration does not necessarily have to be zero when the velocity is zero. For example, when a ball is dropped from rest, its velocity is zero at the instant it starts moving, but it has a non-zero acceleration due to gravity.

C: A can be zero when v = 0.
This is false.
While it's true that acceleration can be zero when velocity is zero, it is not necessarily always the case. For instance, a body at rest could start accelerating immediately after the velocity becomes zero, so it may have zero velocity but a non-zero acceleration.

D: The direction of A must have some correlation with the direction of v.
This is false.
Acceleration can occur in a direction opposite to the velocity, such as when a body is decelerating. For example, when a car is slowing down, its velocity is in the forward direction, but the acceleration is in the opposite (backward) direction.

Therefore, the correct answer is:

Answer: A
Solution: A can be non-zero when v = 0, and A may be zero when v ≠ 0.

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

So, option (d) is the correct answer.

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

The x-t graph of a particle moving in a straight line shows its position over time.

• The slope of the x-t graph indicates the particle's velocity.
• If the slope is constant, the velocity is constant, leading to zero acceleration.
• A changing slope means the velocity is changing, indicating acceleration.
• The a-t graph reflects these changes. A constant slope in the x-t graph results in zero acceleration, while a varying slope results in non-zero acceleration.

Hence, the correct a-t graph is option D, as it accurately represents the acceleration based on the x-t graph data.

_{Hence, option D is correct.}

The particle comes to rest (velocity v = 0) at two instants during the motion. These instants can be found by analyzing the velocity-time graph:

• At t = 0 s, the velocity is 0, so the particle is at rest.
• Between t = 4 s and t = 6 s, the velocity changes linearly from +10 m/s to -20 m/s. We calculate when the velocity is zero during this interval.

Calculate acceleration during 4 s to 6 s interval:

a = (v_final - v_initial) / (t_final - t_initial) = (-20 - 10) / (6 - 4) = -30 / 2 = -15 m/s²

Use equation v = u + at, where u = 10 m/s at t = 4 s:

Set v = 0 (particle at rest):

0 = 10 + (-15) t'

t' = 10 / 15 = 2/3 seconds ≈ 0.67 seconds after 4 s

So, particle comes to rest at t = 4 + 0.67 = 4.67 s (approximately).

At t = 8 s, the velocity again reaches zero as shown by the graph.

Therefore, the particle comes to rest at t = 0 s, approximately at t = 4.67 s, and at t = 8 s.

Among the options, the particle does NOT come to rest exactly at t = 5 s. Hence, option c) is the incorrect statement.

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

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
 

Area under the graph gives distance
Distance
= 1/2 × 10 × 10 + 10 × 10 + 1/2 × 10 × 10
= 50 + 100 + 50
= 200

Concept:

• Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
• These equations are only valid when the acceleration of the body is constant and they move on a straight line.

There are three equations of motion:

V = u + at

V² = u² + 2 a S

S = ut + 1/2at²

Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

Given that:

Initial velocity (u) = 0

Distance (S) = 20 m

Time (t) = 4 sec

Use S = ut + 1/2at²

20 = 0 + 1/2 × a × 4²

Acceleration = a = 20/8 = 2.5 m/s²

Equations of Motion
The equations of motion establish the relationship between acceleration, time, distance, initial speed, and final speed for a body moving in a straight line with uniform acceleration.

The equations are

1. v = u + at 
2. s = ut + 1/2at²
3. v² = u² + 2as 

v is final velocity, u  is initial velocity, t is time, a is acceleration, s is the distance travelled.
Calculation:
Given:
v = 50 m/s, a = 2.5 m/s²?
v² = u² + 2as 
50² = 2 × 2.5 × s
s = 500 m