Question

If x=(3++ 2t² + 2t + 7) m
and t = 5sec.
find velocity, acceleration at t = 0, t = 1 sec​

Answer 1

$x=2t^2+2t+7+3\\ \\ \\x=2t^2+2t+10$

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we know that :-

$VELOCITY=\dfrac{\triangle s}{\triangle t}\\ \\ \\ \\VELOCITY=\dfrac{dx}{dt}$

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Further we know :-

$ACCELERATION=\dfrac{\triangle v}{\triangle t}\\ \\ \\ \\ACCELERATION=\dfrac{dv}{dt}$

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Now:-

Applying the derivative:-

$VELOCITY=\dfrac{dx}{dt}\\ \\ \\VELOCITY=\dfrac{d}{dt}(2t^2+2t+10)\\ \\ \\V =\dfrac{d}{dt}(2t^2)+\dfrac{d}{dt}(2t)+\dfrac{d}{dt}(10)\\ \\ \\V=4t+2+0\\ \\ \\V=4t+2$

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Velocity at :-

t = 0 second:-

$V=4t+2\\ \\ \\V=4(0)+2 = 2\:m/sec$

t=1 second:-

$V=4t+2\\ \\ \\V=4(1)+2\\ \\ \\V=6\:m/sec$

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$ACCELERATION=\dfrac{dv}{dt}\\ \\ \\ACCELERATION=\dfrac{d}{dt}(4t+2)\\ \\ \\ACCELERATION=\dfrac{d}{dt}(4t)+\dfrac{d}{dt}(2)\\ \\ \\ACCELERATION=4+0 =4$

Hence:-

Acceleration is independent of Time.

Thus ,

At t=0 and t=1; Acceleration is equal to 4 m/sec^2