NEET Exam > NEET Questions > A particle accelerates from rest at a consta...
Start Learning for Free
A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled is
- a)32 m
- b)16 m
- c)4 m
- d)8 m
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A particle accelerates from rest at a constant rate for some time and...
8 = a t1 and 0 = 8 – a (4 – t1)
8 = 4 a – 8 or a = 4 and t1 = 8/4 = 2 sec
Now, s1 = 0 x 2 + ½ x 4(2)2 or s1 = 8 m
s2 = 8 x 2 - 1/2 x 4 x (2)2 or s2 = 8m
∴ s1 + s2 = 16 m
Free Test
FREE
| Start Free Test |
Community Answer
A particle accelerates from rest at a constant rate for some time and...
To solve this problem, we can use the equations of motion. Let's break down the problem into two parts: the acceleration phase and the deceleration phase.
Acceleration phase:
Given:
Initial velocity (u) = 0 m/sec
Final velocity (v) = 8 m/sec
Time taken (t) = unknown
We can use the equation of motion:
v = u + at
Substituting the values, we get:
8 = 0 + a*t
Simplifying, we have:
8 = at
Deceleration phase:
Given:
Initial velocity (u) = 8 m/sec
Final velocity (v) = 0 m/sec
Time taken (t) = unknown
Using the same equation of motion:
v = u + at
Substituting the values, we get:
0 = 8 + a*t
Simplifying, we have:
at = -8
Total time:
The total time taken is given as 4 seconds. Therefore, the sum of the time taken during the acceleration and deceleration phases is equal to 4 seconds.
t + t = 4
2t = 4
t = 2 seconds
Distance traveled:
To find the distance traveled, we need to calculate the distance covered during the acceleration phase and the deceleration phase, and then sum them up.
Acceleration phase:
Using the equation of motion:
s = ut + 0.5*a*t^2
Substituting the values, we get:
s1 = 0*2 + 0.5*a*2^2
s1 = 2a
Deceleration phase:
Using the same equation of motion:
s = ut + 0.5*a*t^2
Substituting the values, we get:
s2 = 8*2 + 0.5*(-8)*2^2
s2 = 16 - 16
s2 = 0
Therefore, the total distance traveled is the sum of s1 and s2:
s = s1 + s2
s = 2a + 0
s = 2a
Since s = 16 meters (given in option B), we can solve for a:
2a = 16
a = 8
Therefore, the distance traveled is 16 meters, which is given in option B.
Acceleration phase:
Given:
Initial velocity (u) = 0 m/sec
Final velocity (v) = 8 m/sec
Time taken (t) = unknown
We can use the equation of motion:
v = u + at
Substituting the values, we get:
8 = 0 + a*t
Simplifying, we have:
8 = at
Deceleration phase:
Given:
Initial velocity (u) = 8 m/sec
Final velocity (v) = 0 m/sec
Time taken (t) = unknown
Using the same equation of motion:
v = u + at
Substituting the values, we get:
0 = 8 + a*t
Simplifying, we have:
at = -8
Total time:
The total time taken is given as 4 seconds. Therefore, the sum of the time taken during the acceleration and deceleration phases is equal to 4 seconds.
t + t = 4
2t = 4
t = 2 seconds
Distance traveled:
To find the distance traveled, we need to calculate the distance covered during the acceleration phase and the deceleration phase, and then sum them up.
Acceleration phase:
Using the equation of motion:
s = ut + 0.5*a*t^2
Substituting the values, we get:
s1 = 0*2 + 0.5*a*2^2
s1 = 2a
Deceleration phase:
Using the same equation of motion:
s = ut + 0.5*a*t^2
Substituting the values, we get:
s2 = 8*2 + 0.5*(-8)*2^2
s2 = 16 - 16
s2 = 0
Therefore, the total distance traveled is the sum of s1 and s2:
s = s1 + s2
s = 2a + 0
s = 2a
Since s = 16 meters (given in option B), we can solve for a:
2a = 16
a = 8
Therefore, the distance traveled is 16 meters, which is given in option B.
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer?
Question Description
A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer?.
A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled isa)32 mb)16 mc)4 md)8 mCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup