Question

Please provide correct solution :​

Answer 1

Answer:

Explanation:

Hope this will be helpful to you and please mark my answer as the brainleast answer.plz plz

Answer 2

Hey !

Given,

$\mathsf{v = 2t^{2} e^{-t}}$. ----------[1]

To find:

value of t when Acceleration = 0

Acceleration is the rate of change of Velocity.

$a = \dfrac{dv}{dt}$

Substitute [1]

$\implies a = \dfrac{d}{dt}\, (2t^{2} e^{-t})$

$\implies a =2\left[\dfrac{d}{dt}\, (t^{2} e^{-t})\right]$

From uv rule, we have,

$\huge{\fbox{\mathbf{d(uv) = vu ' + uv'}}}$

$\implies a =2\left[\dfrac{d}{dt}\, (t^{2} \: e^{-t})\right]$

Apply uv rule,

$\implies a =2\left[(t^{2}\, e^{-t} \cdot (-1))+ (2t \, e^{-t})\right]$

$\implies a =2\left[(t^{2}\, e^{-t} \cdot (-1))+ (2t \, e^{-t})\right] \\ \\ \implies \:a =2\left[2t \, e^{-t} -t^{2}\, e^{-t} \right] \: \\ \\ \implies \:a =2te^{-t}\left[2 -t \right]$

Now,

Equate a to 0

$\implies 2te^{-t}[2-t] = 0$

Divide by 2t on both sides

$\implies e^{-t}[2-t] = 0$

Here, we have $e^{-t} = 0$, which is really not possible as its value will be 0.00..... , but not exactly zero for all values except 0.

when t =0, $e^{0}$, becomes 1.

And, we have

$\implies 2-t = 0$

$\implies t = 2$

Therefore, Acceleration is zero at t = 2 seconds as per the velocity given.





Option - (4) is the answer.

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