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A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?
- a)1
- b)2
- c)3
- d)4
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A particle is moving in a straight line and its distance x from a fixe...
Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1
So, v = dx/dt = t3/3 – 2t2/ + 3t + 1
Thus, dv/dt = t2 – 4t + 3
And d2v/dt2 = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t2 – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1
So, v = dx/dt = t3/3 – 2t2/ + 3t + 1
Thus, dv/dt = t2 – 4t + 3
And d2v/dt2 = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t2 – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d2v/dt2]t = 3 = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
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Community Answer
A particle is moving in a straight line and its distance x from a fixe...
Understanding the Problem
The position of the particle is given by the equation:
x = (t^4)/12 - (2t^3)/3 + (3t^2)/2 + t + 15.
To find the time at which the velocity is minimum, we first need to determine the velocity of the particle.
Finding Velocity
- The velocity (v) is the derivative of the position function (x) with respect to time (t).
So, we differentiate the position function:
v = dx/dt = (1/12)(4t^3) - (2/3)(3t^2) + (3/2)(2t) + 1.
This simplifies to:
v = (t^3)/3 - 2t^2 + 3t + 1.
Finding Minimum Velocity
- To find the minimum velocity, we need to set the first derivative of the velocity function (acceleration) to zero:
a = dv/dt.
This involves differentiating the velocity function:
a = (1/3)(3t^2) - 4t + 3.
This simplifies to:
a = t^2 - 4t + 3.
Next, we set this equal to zero to find critical points:
t^2 - 4t + 3 = 0.
Solving the Quadratic Equation
- Factoring gives:
(t - 3)(t - 1) = 0.
Thus, t = 1 and t = 3.
Determining Minimum Velocity
- We need to check the second derivative (d^2v/dt^2) or the nature of critical points to confirm which is a minimum:
- Testing intervals around t = 1 and t = 3 shows that at t = 3, the velocity is minimum.
Conclusion
- Therefore, the time at which the velocity is minimum is t = 3 seconds, confirming option C as the correct answer.
The position of the particle is given by the equation:
x = (t^4)/12 - (2t^3)/3 + (3t^2)/2 + t + 15.
To find the time at which the velocity is minimum, we first need to determine the velocity of the particle.
Finding Velocity
- The velocity (v) is the derivative of the position function (x) with respect to time (t).
So, we differentiate the position function:
v = dx/dt = (1/12)(4t^3) - (2/3)(3t^2) + (3/2)(2t) + 1.
This simplifies to:
v = (t^3)/3 - 2t^2 + 3t + 1.
Finding Minimum Velocity
- To find the minimum velocity, we need to set the first derivative of the velocity function (acceleration) to zero:
a = dv/dt.
This involves differentiating the velocity function:
a = (1/3)(3t^2) - 4t + 3.
This simplifies to:
a = t^2 - 4t + 3.
Next, we set this equal to zero to find critical points:
t^2 - 4t + 3 = 0.
Solving the Quadratic Equation
- Factoring gives:
(t - 3)(t - 1) = 0.
Thus, t = 1 and t = 3.
Determining Minimum Velocity
- We need to check the second derivative (d^2v/dt^2) or the nature of critical points to confirm which is a minimum:
- Testing intervals around t = 1 and t = 3 shows that at t = 3, the velocity is minimum.
Conclusion
- Therefore, the time at which the velocity is minimum is t = 3 seconds, confirming option C as the correct answer.
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A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer?.
A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t4/12 – 2t3/3 + 3t2/2 + t + 15. At what time is the velocity minimum?a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer?.
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