TS EAMCET · Physics · Motion In One Dimension

The velocity of a particle moving along the $x$ - axis varies as a function of time $t$ as $v (t) = (1 - 3 t^{2} + 2 t^{3}) m s^{- 1}$ If its position at $t = 0$ is $x = 0$ then at $= 2 s,$ it's position is

A $1 m$
B $2 m$
C $4 m$
D $18 m$

Verified Solution

Answer & Solution

Correct Answer

(B) $2 m$

Step-by-step Solution

Detailed explanation

The given function of velocity is, vt=1-3t2+2t3 m s-1, we know that the rate of change of displacement is called velocity, now the displacement can be calculated by integrating it, ⇒v=dxdt⇒∫dx=∫vdt⇒x=t-t3+t42 at, t=2 s, the position…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The velocity of a particle moving along the x- axis varies as a function of time t as vt=1-3t2+2t3 m s-1 If its position at t=0 is x=0 then at =2 s, it's position is

Step-by-step Solution

The velocity of a particle moving along the $x$ - axis varies as a function of time $t$ as $v (t) = (1 - 3 t^{2} + 2 t^{3}) m s^{- 1}$ If its position at $t = 0$ is $x = 0$ then at $= 2 s,$ it's position is