NEET UG PHYSICS MCQ: MOTION IN A STRAIGHT LINE

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Analyzing +4 | -1

A train accelerates from rest at a constant rate α for distance

{x_1}

and time

{t_1}

. After that it retards to rest at constant rate

\beta

for distance

{x_2}

and time

{t_2}

. Then it is found that

View answer and explanation

Correct option: B

\frac{{{x_1}}}{{{x_2}}} = \frac{\beta }{\alpha } = \frac{{{t_1}}}{{{t_2}}}

Option C (

\frac{{{x_1}}}{{{x_2}}} = \frac{\beta }{\alpha } = \frac{{{t_2}}}{{{t_1}}}

) is correct because it relates the distances, accelerations, and times during acceleration and retardation.

More Questions from MOTION IN A STRAIGHT LINE

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8 \mathrm{~m} / \mathrm{s}

30\mathrm{km}{\mathrm{h}^{ - 1}}

1 \mathrm{~ms}^{-2}

150 \mathrm{~m}

2 \mathrm{~ms}^{-2}

{V_1} = 10\,kmph

{V_2} = 20\,kmph

{V_3} = 60\,kmph