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. What is the minimum velocity?a)1 cm/secb)2 cm/secc)3 cm/secd)4 cm/secCorrect answer is option 'A'. Can you explain this answer?

Question

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 &ndash; 2t3/3 + 3t2/2 + t + 15. What is the minimum velocity?a)1 cm/secb)2 cm/secc)3 cm/secd)4 cm/secCorrect answer is option 'A'. Can you explain this answer?

Answer

The equation x = t^4/12 represents the motion of the particle as it moves in a straight line. The distance x from a fixed point on the line at any time t seconds is given by this equation.

To understand the meaning of this equation, let's break it down:

- The variable x represents the distance of the particle from a fixed point on the line. It can be measured in any unit of length (e.g., meters, feet).
- The variable t represents time, measured in seconds.
- The equation x = t^4/12 indicates that the distance x is equal to t raised to the power of 4, divided by 12.

This equation suggests that the distance of the particle from the fixed point increases as time passes, and the rate of increase is proportional to the fourth power of time. The constant factor of 1/12 determines the scale of this relationship.

For example, if t = 2 seconds, then x = (2^4)/12 = 16/12 = 4/3. This means that at 2 seconds, the particle is located 4/3 units away from the fixed point.

Similarly, if t = 3 seconds, then x = (3^4)/12 = 81/12 = 6.75. This means that at 3 seconds, the particle is located approximately 6.75 units away from the fixed point.

In summary, the equation x = t^4/12 describes the distance x of a particle from a fixed point on a straight line as a function of time t.

Answer

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.
Putting t = 3 in (1) we get,
3³/3 – 2(3)²/ + 3(3) + 1
= 1 cm/sec.

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