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Standard IX

Physics

Derivation of Position-Velocity Relation by Graphical Method

An object sta...

Question

An object starting from rest moves on a straight road for time t and comes to rest finally. The distance is converted in two parts. In the first part it is accelerated at constant acceleration $α$ and then decelerated at rate $β$ . Then maximum velocity is

$α t$

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$β t$

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$(\frac{α + β}{2}) t$

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$(\frac{α β}{α + β}) t$

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Solution

The correct option is D
$(\frac{α β}{α + β}) t$

$V = (0) + α t_{1} . . . . . . . . (i) 0 = (α t_{1}) + β t_{2} . . . . . . . (i i) t_{1} + t_{2} = t \Rightarrow V = (\frac{α β}{α + β}) (t)$

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An object starting from rest moves on a straight road for time t and comes to rest finally. The distance is converted in two parts. In the first part it is accelerated at constant acceleration α and then decelerated at rate β. Then maximum velocity is

The correct option is D (αβα+β)t V=(0)+αt1 ........(i)0=(αt1)+βt2 .......(ii)t1+t2=t⇒V=(αβα+β)(t)

An object starting from rest moves on a straight road for time t and comes to rest finally. The distance is converted in two parts. In the first part it is accelerated at constant acceleration $α$ and then decelerated at rate $β$ . Then maximum velocity is

The correct option is D
$(\frac{α β}{α + β}) t$

$V = (0) + α t_{1} . . . . . . . . (i) 0 = (α t_{1}) + β t_{2} . . . . . . . (i i) t_{1} + t_{2} = t \Rightarrow V = (\frac{α β}{α + β}) (t)$