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A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, are
- a)a = 50 m/s2; d = 125 m
- b)a = 100 m/s2: d = 100 m
- c)ci = 100 m/s2: d = 125 m
- d)a = 50 m/s2; d = 100 m
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A particle starts from rest with a constant acceleration. At a time t ...
Velocity at time t is
100 = a. t ...(i)
and velocity at time (t + 1) is
150 = a (t + 1) ...(ii)
Subtracting (i) from (ii). a = 50 m/s2
During the interval t and (t + 1) second
time elapsed = 1 s
intial velocity u = 100 m/s
final velocity v = 150 m/s
100 = a. t ...(i)
and velocity at time (t + 1) is
150 = a (t + 1) ...(ii)
Subtracting (i) from (ii). a = 50 m/s2
During the interval t and (t + 1) second
time elapsed = 1 s
intial velocity u = 100 m/s
final velocity v = 150 m/s
Most Upvoted Answer
A particle starts from rest with a constant acceleration. At a time t ...
To solve this question, we can use the equations of motion. Let's break down the problem into smaller parts:
1. Initial conditions:
- The particle starts from rest, which means its initial velocity (u) is 0 m/s.
2. Speed at t seconds:
- At time t, the speed of the particle is given as 100 m/s.
3. Speed at (t+1) seconds:
- One second later, at (t+1) seconds, the speed of the particle is given as 150 m/s.
4. Finding acceleration:
- We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
- Using the information given, we can substitute the values:
- At t seconds: 100 = 0 + a * t
- At (t+1) seconds: 150 = 0 + a * (t+1)
- Subtracting the first equation from the second equation, we get:
- 150 - 100 = a * (t+1) - a * t
- 50 = a * 1
- Therefore, the acceleration (a) is 50 m/s^2.
5. Finding distance travelled in (t+1)th second:
- We can use the equation v = u + at to find the distance travelled in the (t+1)th second.
- Since the particle starts from rest, the initial velocity (u) is 0 m/s.
- Plugging in the values, we have:
- 150 = 0 + 50 * (t+1)
- 150 = 50t + 50
- 100 = 50t
- t = 2 seconds
- Therefore, the distance travelled in the (t+1)th second is:
- d = ut + (1/2)at^2
- d = 0 + (1/2) * 50 * (2)^2
- d = 0 + 50 * 4
- d = 200 m
So, the correct answer is option A:
a = 50 m/s^2 and d = 125 m.
1. Initial conditions:
- The particle starts from rest, which means its initial velocity (u) is 0 m/s.
2. Speed at t seconds:
- At time t, the speed of the particle is given as 100 m/s.
3. Speed at (t+1) seconds:
- One second later, at (t+1) seconds, the speed of the particle is given as 150 m/s.
4. Finding acceleration:
- We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
- Using the information given, we can substitute the values:
- At t seconds: 100 = 0 + a * t
- At (t+1) seconds: 150 = 0 + a * (t+1)
- Subtracting the first equation from the second equation, we get:
- 150 - 100 = a * (t+1) - a * t
- 50 = a * 1
- Therefore, the acceleration (a) is 50 m/s^2.
5. Finding distance travelled in (t+1)th second:
- We can use the equation v = u + at to find the distance travelled in the (t+1)th second.
- Since the particle starts from rest, the initial velocity (u) is 0 m/s.
- Plugging in the values, we have:
- 150 = 0 + 50 * (t+1)
- 150 = 50t + 50
- 100 = 50t
- t = 2 seconds
- Therefore, the distance travelled in the (t+1)th second is:
- d = ut + (1/2)at^2
- d = 0 + (1/2) * 50 * (2)^2
- d = 0 + 50 * 4
- d = 200 m
So, the correct answer is option A:
a = 50 m/s^2 and d = 125 m.
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A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer?
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A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer?.
A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer?.
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A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. The acceleration a and the distance d travelled during the (t + 1)th second respectively, area)a = 50 m/s2; d = 125 mb)a = 100 m/s2: d = 100 mc)ci = 100 m/s2: d = 125 md)a = 50 m/s2;d = 100 mCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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