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[SOLVED] Velocity and acceleration vectors and their magnitudes

Dhamnekar Winod

Active member
Nov 17, 2018
163
1626971733313.png

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

DaalChawal

Member
Apr 8, 2021
89
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
Nov 17, 2018
163
$v=\sqrt{37}, a= \sqrt{325}$ at t=0
 

DaalChawal

Member
Apr 8, 2021
89
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
Nov 17, 2018
163
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
Apr 8, 2021
89
$x= e^{-t}$

$v_x = - e^{-t}$

$a_x = e^{-t}$
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
Nov 17, 2018
163
$x= e^{-t}$

$v_x = - e^{-t}$

$a_x = e^t$
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
Apr 8, 2021
89
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
Jan 30, 2018
821
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}