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Instantaneous Speed

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Question

Starting from rest, a particle is accelerated for time $t_{1}$ with constant acceleration $a_{1}$ and then stopped in time $t_{2}$ with constant deceleration $a_{2}$ . Let $v_{1}$ be the average velocity in this case and $s_{1}$ the total displacement. In the second case, the particle is accelerated for the same time $t_{1}$ with constant acceleration $2 a_{1}$ and brought to rest with constant retardation $a_{2}$ in time $t_{3}$ . If $v_{2}$ is the average velocity in this case and $s_{2}$ the total displacement, then

2 v_{1} < v_{2} < 4 v_{1}

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None

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v_{2} = 2 v_{1}

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s_{2} = 2 s_{1}

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Solution

The correct option is B None
For $v_{1}$ :

From figure
$v_{1} = \frac{Total displacement}{Total time}$

$= \frac{Area under v-t curve}{Total time}$

$v_{1} = \frac{s_{1}}{t_{1} + t_{2}}$

$= \frac{1}{2} (a_{1} t_{1}) (t_{1} + t_{2}) (\frac{1}{t_{1} + t_{2}}$ )

$v_{1} = \frac{1}{2} a_{1} t_{1} . . . (i)$

Similarly, for $v_{2}$ :

$s_{2} = \frac{1}{2} (2 a_{1} t_{1}) (t_{1} + t_{3})$

$v_{2} = \frac{s_{2}}{t_{1} + t_{2}}$

$= \frac{1}{2} (2 a_{1} t_{1}) (t_{1} + t_{3}) (\frac{1}{t_{1} + t_{3}})$

$v_{2} = a_{1} t_{1}$

From (𝑖) and (𝑖𝑖)

$v_{2} = 2 v_{1}$

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Instantaneous Speed

Standard IX Physics

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The correct option is B NoneFor v1: From figure v1=Total displacementTotal time =Area under v-t curveTotal time v1=s1t1+t2 =12(a1t1)(t1+t2)(1t1+t2) v1=12a1t1...(i) Similarly, for v2: s2=12(2a1t1)(t1+t3) v2=s2t1+t2 =12(2a1t1)(t1+t3)(1t1+t3) v2=a1t1 From (𝑖) and (𝑖𝑖) v2=2v1