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Instantaneous acceleration

The graph sho...

Question

The graph shows the variation of $\frac{1}{v}$ (where $v$ is the velocity of the particle) with respect to time. Then find the value of acceleration at $t = 3 sec$

3 {m/s}^{2}

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5 {m/s}^{2}

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1 {m/s}^{2}

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0.5 {m/s}^{2}

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Solution

The correct option is A $3 {m/s}^{2}$
Slope $= \frac{d (\frac{1}{v})}{d t} = - 1$
$\Rightarrow - \frac{1}{v^{2}} \frac{d v}{d t} = - 1$
$\Rightarrow {(\frac{1}{\sqrt{3}})}^{2} \frac{d v}{d t} = 1$
$\Rightarrow a = \frac{d v}{d t} = 3 {m/s}^{2}$

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Figure shows the acceleration – time ( $a - t$ ) graph of a body moving in a straight line. Which graph in figure shows the velocity – time ( $v - t$ ) graph of the motion of the body? (Assume that $x = 0$ and $v = 0$ at $t = 0$ )

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The graph shows the variation of 1v (where v is the velocity of the particle) with respect to time. Then find the value of acceleration at t=3 sec

The correct option is A 3 m/s2Slope =d(1v)dt=−1 ⇒−1v2dvdt=−1 ⇒(1√3)2dvdt=1 ⇒a=dvdt=3 m/s2

The graph shows the variation of $\frac{1}{v}$ (where $v$ is the velocity of the particle) with respect to time. Then find the value of acceleration at $t = 3 sec$

The correct option is A $3 {m/s}^{2}$
Slope $= \frac{d (\frac{1}{v})}{d t} = - 1$
$\Rightarrow - \frac{1}{v^{2}} \frac{d v}{d t} = - 1$
$\Rightarrow {(\frac{1}{\sqrt{3}})}^{2} \frac{d v}{d t} = 1$
$\Rightarrow a = \frac{d v}{d t} = 3 {m/s}^{2}$