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A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds,respectively.What is the distance described by the particle in 3 seconds?
- a)30 cm
- b)31 cm
- c)32 cm
- d)33 cm
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A particle moves with uniform acceleration along a straight line and d...
Given data:
- Distance at 2 seconds: 21m
- Distance at 4 seconds: 43m
- Distance at 7 seconds: 91m
Calculating acceleration:
- Average acceleration = Change in velocity / Change in time
- Change in velocity = (Final velocity - Initial velocity)
- Change in time = (Final time - Initial time)
Using the given data:
- Initial time (t1) = 2 seconds
- Final time (t2) = 4 seconds
- Initial distance (d1) = 21m
- Final distance (d2) = 43m
Change in velocity = (43 - 21) / (4 - 2) = 22 / 2 = 11 m/s^2
Calculating final velocity:
- Final velocity = Initial velocity + (Acceleration * Time)
- Initial velocity = (Final velocity - (Acceleration * Time))
At t = 4 seconds:
- Initial velocity = 43 - (11 * 2) = 43 - 22 = 21 m/s
At t = 7 seconds:
- Distance = 21 + (21 * 3) + (0.5 * 11 * 3^2)
- Distance = 21 + 63 + 49.5 = 133.5 m
Therefore, the distance described by the particle in 3 seconds is 133.5m.
- Distance at 2 seconds: 21m
- Distance at 4 seconds: 43m
- Distance at 7 seconds: 91m
Calculating acceleration:
- Average acceleration = Change in velocity / Change in time
- Change in velocity = (Final velocity - Initial velocity)
- Change in time = (Final time - Initial time)
Using the given data:
- Initial time (t1) = 2 seconds
- Final time (t2) = 4 seconds
- Initial distance (d1) = 21m
- Final distance (d2) = 43m
Change in velocity = (43 - 21) / (4 - 2) = 22 / 2 = 11 m/s^2
Calculating final velocity:
- Final velocity = Initial velocity + (Acceleration * Time)
- Initial velocity = (Final velocity - (Acceleration * Time))
At t = 4 seconds:
- Initial velocity = 43 - (11 * 2) = 43 - 22 = 21 m/s
At t = 7 seconds:
- Distance = 21 + (21 * 3) + (0.5 * 11 * 3^2)
- Distance = 21 + 63 + 49.5 = 133.5 m
Therefore, the distance described by the particle in 3 seconds is 133.5m.
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Community Answer
A particle moves with uniform acceleration along a straight line and d...
We assume that the particle moves with uniform acceleration 2f m/sec.
Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.
Let, v be the velocity of the particle at time t seconds, then,
So, dv/dt = 2f
Or ∫dv = ∫2f dt
Or v = 2ft + b ……….(1)
Or dx/dt = 2ft + b
Or ∫dx = 2f∫tdt + ∫b dt
Or x = ft2 + bt + a ……….(2)
Where, a and b are constants of integration.
Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.
Putting these values in (2) we get,
4f + 2b + a = 21 ……….(3)
16f + 4b + a = 43 ……….(4)
49f + 7b + a = 91 ……….(5)
Solving (3), (4) and (5) we get,
a = 7, b = 5 and f = 1
Therefore, from (2) we get,
x = t2 + 5t + 7
Therefore, the distance described by the particle in 3 seconds,
= [x]t = 3 = (32 + 5*3 + 7)m = 31m
Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.
Let, v be the velocity of the particle at time t seconds, then,
So, dv/dt = 2f
Or ∫dv = ∫2f dt
Or v = 2ft + b ……….(1)
Or dx/dt = 2ft + b
Or ∫dx = 2f∫tdt + ∫b dt
Or x = ft2 + bt + a ……….(2)
Where, a and b are constants of integration.
Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.
Putting these values in (2) we get,
4f + 2b + a = 21 ……….(3)
16f + 4b + a = 43 ……….(4)
49f + 7b + a = 91 ……….(5)
Solving (3), (4) and (5) we get,
a = 7, b = 5 and f = 1
Therefore, from (2) we get,
x = t2 + 5t + 7
Therefore, the distance described by the particle in 3 seconds,
= [x]t = 3 = (32 + 5*3 + 7)m = 31m
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A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds,respectively.What is the distance described by the particle in 3 seconds?a)30 cmb)31 cmc)32 cmd)33 cmCorrect answer is option 'B'. Can you explain this answer?
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A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds,respectively.What is the distance described by the particle in 3 seconds?a)30 cmb)31 cmc)32 cmd)33 cmCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds,respectively.What is the distance described by the particle in 3 seconds?a)30 cmb)31 cmc)32 cmd)33 cmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds,respectively.What is the distance described by the particle in 3 seconds?a)30 cmb)31 cmc)32 cmd)33 cmCorrect answer is option 'B'. Can you explain this answer?.
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