AP EAMCET · PHYSICS · Motion In One Dimension

A car moving with a velocity $6.25 \mathrm{~ms}^{-1}$ is decelerated with $2.5 \sqrt{\mathrm{v}} \mathrm{ms}^{-2}$ ( $v$ is instantaneous velocity). Time taken by the car to come to rest is

A $2 \mathrm{~s}$
B $3 \mathrm{~s}$
C $2.5 \mathrm{~s}$
D $4 \mathrm{~s}$

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Answer & Solution

Correct Answer

(A) $2 \mathrm{~s}$

Step-by-step Solution

Detailed explanation

Given, $u=6.25 \mathrm{~ms}^{-1}$ and $a=-2.5 \sqrt{v} \mathrm{~ms}^{-2}$, $v=$ instantaneous velocity As we know, $a=\frac{d v}{d t}$ So, $\frac{d v}{d t}=-2.5 \sqrt{v}$ Integrate on the both sides, we get…

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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A body is projected horizontally from the top of a tower of height

180 m

with a velocity of

20 {ms}^{- 1}

. If acceleration due to gravity is

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then match the following.

List-I		List-II
$(A)$	Velocity of the body after $1$ second $($ in ${ms}^{- 1}$ )	(I)	$5$
$(B)$	Horizontal displacement of the body after $1$ second (in meters)	(ii)	$20$
$(C)$	Vertical displacement of the body after $1$ second (in meters)	(iii)	$10$
$(D)$	Vertical velocity of the body after $1$ second $($ in ${ms}^{- 1}$ )	(iv)	$22.4$

The correct answer is

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A car moving with a velocity \(6.25 \mathrm{~ms}^{-1}\) is decelerated with \(2.5 \sqrt{\mathrm{v}} \mathrm{ms}^{-2}\) ( \(v\) is instantaneous velocity). Time taken by the car to come to rest is

Answer & Solution

Step-by-step Solution

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