back to directory
NEET PHYSICSMOTION IN A STRAIGHT LINEEasy

Question

A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd second?

A

7/5

B

5/7

C

7/3

D

3/7

Step-by-Step Solution

  1. Method 1: Galileo's Law of Odd Numbers For a particle starting from rest with uniform acceleration, the distances traversed during equal successive intervals of time are in the ratio of odd numbers: 1:3:5:7:91 : 3 : 5 : 7 : 9 \dots . Distance in 1st second 1\propto 1 Distance in 2nd second 3\propto 3 Distance in 3rd second 5\propto 5 Distance in 4th second 7\propto 7 Therefore, the ratio of the distance travelled in the 4th second to that in the 3rd second is 7:57:5 or 7/57/5.
  2. Method 2: Kinematic Formula Distance travelled in the nn-th second is given by Sn=u+a2(2n1)S_n = u + \frac{a}{2}(2n - 1). Since the body starts from rest, u=0u = 0. S4=a2(2×41)=7a2S_4 = \frac{a}{2}(2 \times 4 - 1) = \frac{7a}{2}. S3=a2(2×31)=5a2S_3 = \frac{a}{2}(2 \times 3 - 1) = \frac{5a}{2}.
  • Ratio S4S3=7a/25a/2=75\frac{S_4}{S_3} = \frac{7a/2}{5a/2} = \frac{7}{5}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A STRAIGHT LINE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A STRAIGHT LINEstartsdistancetravelledduringsecond

More MOTION IN A STRAIGHT LINE Questions

View all

A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the ball during its height if the air resistance is not ignored?

A.Option 1
B.Option 2
C.4
D.Option 4
MediumSolve

The velocity of a bullet is reduced from $200 \text{ m/s}$ to $100 \text{ m/s}$ while travelling through a wooden block of thickness $10 \text{ cm}$. The retardation, assuming it to be uniform, will be:

A.$10 \times 10^4 \text{ m/s}^2$
B.$12 \times 10^4 \text{ m/s}^2$
C.$13.5 \times 10^4 \text{ m/s}^2$
D.$15 \times 10^4 \text{ m/s}^2$
MediumSolve

A body A starts from rest with an acceleration $a_1$. After $2$ seconds, another body B starts from rest with an acceleration $a_2$. If they travel equal distances in the $5^{\text{th}}$ second, after the start of A, then the ratio $a_1 : a_2$ is equal to:

A.5 : 9
B.5 : 7
C.9 : 5
D.9 : 7
MediumSolve

The displacement of a particle, moving in a straight line, is given by $s = 2t^2 + 2t + 4$ where $s$ is in metres and $t$ in seconds. The acceleration of the particle is:

A.2 m/s²
B.4 m/s²
C.6 m/s²
D.8 m/s²
EasySolve

A car moving with a velocity of $10 \text{ m/s}$ can be stopped by the application of a constant force $F$ in a distance of $20 \text{ m}$. If the velocity of the car is $30 \text{ m/s}$, it can be stopped by this force in:

A.20/3 m
B.20 m
C.60 m
D.180 m
MediumSolve

Which of the following velocity-time graphs shows a realistic situation for a body in motion?

A.Option 1
B.Option 2
C.Option 3
D.Option 4
EasySolve

Time taken by an object falling from rest to cover the height of $h_1$ and $h_2$ is respectively $t_1$ and $t_2$. Then the ratio of $t_1$ to $t_2$ is:

A.$h_1 : h_2$
B.$\sqrt{h_1} : \sqrt{h_2}$
C.$h_1 : 2h_2$
D.$2h_1 : h_2$
EasySolve

A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is:

A.4.9 m
B.9.8 m
C.19.6 m
D.24.5 m
MediumSolve

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →