A particle travels in a straight …
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Arun Soni 6 years, 9 months ago
{tex}v = \frac{4}{a}{/tex}(Given)
{tex}\therefore a = \frac{{dv}}{{dt}} = \frac{4}{v}{/tex} -eqn (1)
{tex}\Rightarrow\int\limits_6^v {vdv} = \int\limits_2^t {4dt} {/tex}
{tex} \Rightarrow \frac{1}{2}{v^2} - 18 = 4t - 8{/tex}
{tex} \Rightarrow {v^2} = 18t + 20{/tex}
At t=3 sec,
v2=74
{tex}v = \pm \sqrt {74} {/tex}
Taking positive value of v
{tex}v = +6.63m/\sec {/tex}
Putting value of v in eq n (1)
{tex}a = \frac{4}{{6.63}} = 0.603m/{\sec ^2}{/tex}
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