# [SOLVED]Velocity and acceleration vectors and their magnitudes

#### Dhamnekar Winod

##### Active member

How to answer this question? I am working on this question. Any math help, hint or even correct answer will be accepted.

#### DaalChawal

##### Member
Differentiate them to get velocities and accelerations in x, y, z(i, j, k ) directions. Add them vectorially to get resultant velocity and acceleration.

#### Dhamnekar Winod

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

#### DaalChawal

##### Member
so? what problem are you facing? Once you get velocity and acceleration, calculate the magnitude by $\sqrt[]{a^2 + b^2 +c^2}$ where a vector is given by $a i + b j + c k$

#### Dhamnekar Winod

##### Active member
so? what problem are you facing? Once you get velocity and acceleration, calculate the magnitude by $\sqrt[]{a^2 + b^2 +c^2}$ where a vector is given by $a i + b j + c k$
Answers given are magnitudes of velocity and acceleration vectors at t=0. What are you talking about?

#### DaalChawal

##### Member
$x= e^{-t}$

$v_x = - e^{-t}$

$a_x = e^{-t}$
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

#### Dhamnekar Winod

##### Active member
$x= e^{-t}$

$v_x = - e^{-t}$

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

#### DaalChawal

##### Member
Okay then use formula that if a vector is expressed as $x = a i + b j + c k$ then its magnitude is given by $|x| = \sqrt[]{a^2+b^2+c^2}$

#### Country Boy

##### Well-known member
MHB Math Helper
Since this has been sitting a while, yes, the answers given in post #3 are corret.

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}