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#### Dhamnekar Winod

##### Active member

- Nov 17, 2018

- 163

- Thread starter Dhamnekar Winod
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- #1

- Nov 17, 2018

- 163

- Apr 8, 2021

- 89

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- #3

- Nov 17, 2018

- 163

$v=\sqrt{37}, a= \sqrt{325}$ at t=0

- Apr 8, 2021

- 89

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- #5

- Nov 17, 2018

- 163

Answers given are magnitudes of velocity and acceleration vectors at t=0. What are you talking about?

- Apr 8, 2021

- 89

$v_x = - e^{-t}$

$a_x = e^{-t}$

similarly calculate velocity and acceleration in $y, z$ directions.

Then

You will get velocity as $- e^{-t} i +(-6)sin(3t) j + 6cos(3t) k$

Now put t=0 to get velocity at t=0nand calculate the magnitude.

for acceleration try yourself

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- #7

- Nov 17, 2018

- 163

I already computed velocity and acceleration vectors. But i only posted here their magnitudes.$x= e^{-t}$

$v_x = - e^{-t}$

$a_x = e^t$

similarly calculate velocity and acceleration in $y, z$ directions.

Then

You will get velocity as $- e^{-t} i +(-6)sin(3t) j + 6cos(3t) k$

Now put t=0 to get velocity at t=0nand calculate the magnitude.

for acceleration try yourself

- Apr 8, 2021

- 89

- Jan 30, 2018

- 821

We have x= e^{-t}, y= 2 cos(3t), and z= 2 sin(3t).

The velocity is given by x'= -e^{-t}, y'= -6 sin(3t), and z'= 6 cos(3t).

The acceleration is given by x''= e^{-t}, y''= -18 cos(3t), and z''= -18 sin(3t).

|v(0)|= sqrt{1+ 0+ 36}= sqrt{37}

|a(0)|= sqrt{1+ 324+ 0}= sqrt{325}