Question

This question has multiple correct options

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Solution

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Correct options are A) and B)

From $t=A$ to $t=B$ velocity increases from v to 0 where v is the velocity at $t=A$

Hence, $v_{avg}$ during interval 0 to A is same as $v_{avg}$ during interval A to B

$A−0v_{c}−0 =tan60_{∘}=3 $;

$∴v_{c}=3 ×OA$

$A−0v_{c}−0 =tan150_{∘}=−tan30_{∘}=−3 1 $

$∴v_{c}=3 1 ×AB$

Equating both $v_{c}$, $3 ×OA=3 1 ×AB$

$⇒OAAB =3$

$⇒OA+ABOA =1+31 $

$⇒OBOA =41 $

Distance covered on in time OA: $d_{OA}=21 ×OA×v_{A}$

Distance covered on in time AB: $d_{AB}=21 ×AB×v_{A}$

$∣a_{OA}∣=tan60_{∘}=3 $

$∣a_{AB}∣=tan30_{∘}3 1 $

$∣a_{AB}∣∣a_{OA}∣ =3$

$∴d_{AB}d_{OA} =21 ×AB×v_{A}21 ×OA×v_{A} $

$∴d_{AB}d_{OA} =cot30AC cot60AC =tan60_{∘}×cot30_{∘}=3$

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