Mixing
At what time $t$ is velocity perpendicular to acceleration?
$begingroup$
So, for $textbf{r}=textbf{r}(t)$,
$textbf{r}=(t+3)textbf{i}+(2t-t^{2})textbf{j}+(3t-t^{2})textbf{k}$ = displacement
Therefore,
$textbf{v}=textbf{r}'=textbf{i}+2textbf{j}-2ttextbf{j}+3textbf{k}-2ttextbf{k}$ = velocity
And,
$textbf{a}=textbf{r}''=-2textbf{j}-2textbf{k}$ = acceleration
To show orthogonality, I know $textbf{a} cdot textbf{b}=0$, so for velocity and acceleration to be orthogonal (or perpendicular), $textbf{v} cdot textbf{a}=0$.
$textbf{v} cdot textbf{a}= (textbf{i}+2textbf{j}-2ttextbf{j}+3textbf{k}-2ttextbf{k}) cdot (-2textbf{j}-2textbf{k})=[textbf{i}+(2+2t)textbf{j}+(3-2t)textbf{k}] cdot (-2textbf{j}-2textbf{k})=(1)(0)textbf{i}+(2+2t)(-2)textbf{j}+(3-2t)(-2)textbf{k}=-4textbf{j}-4ttextbf{j}-6textbf{k}+4ttextbf{k}=0$
$implies 4ttextbf{j}-4ttextbf{k}=-4textbf{j}-6textbf{k}$
Obviously, there is no value for $t$ which satisfies the above, so I think I may have made a mistake!
Thanks.
vectors orthogonality classical-mechanics
asked May 2 '17 at 6:32
Kyle JohnsonKyle Johnson
275
$endgroup$
add a comment |
$begingroup$
So, for $textbf{r}=textbf{r}(t)$,
$textbf{r}=(t+3)textbf{i}+(2t-t^{2})textbf{j}+(3t-t^{2})textbf{k}$ = displacement
Therefore,
$textbf{v}=textbf{r}'=textbf{i}+2textbf{j}-2ttextbf{j}+3textbf{k}-2ttextbf{k}$ = velocity
And,
$textbf{a}=textbf{r}''=-2textbf{j}-2textbf{k}$ = acceleration
To show orthogonality, I know $textbf{a} cdot textbf{b}=0$, so for velocity and acceleration to be orthogonal (or perpendicular), $textbf{v} cdot textbf{a}=0$.
$textbf{v} cdot textbf{a}= (textbf{i}+2textbf{j}-2ttextbf{j}+3textbf{k}-2ttextbf{k}) cdot (-2textbf{j}-2textbf{k})=[textbf{i}+(2+2t)textbf{j}+(3-2t)textbf{k}] cdot (-2textbf{j}-2textbf{k})=(1)(0)textbf{i}+(2+2t)(-2)textbf{j}+(3-2t)(-2)textbf{k}=-4textbf{j}-4ttextbf{j}-6textbf{k}+4ttextbf{k}=0$
$implies 4ttextbf{j}-4ttextbf{k}=-4textbf{j}-6textbf{k}$
Obviously, there is no value for $t$ which satisfies the above, so I think I may have made a mistake!
Thanks.
vectors orthogonality classical-mechanics
asked May 2 '17 at 6:32
Kyle JohnsonKyle Johnson
275
$endgroup$
1
$begingroup$
The dot product if often called the "scalar product". That reminds its users that its output is a scalar not a vector.
$endgroup$
– Lord Shark the Unknown
May 2 '17 at 6:34
add a comment |
$begingroup$
So, for $textbf{r}=textbf{r}(t)$,
$textbf{r}=(t+3)textbf{i}+(2t-t^{2})textbf{j}+(3t-t^{2})textbf{k}$ = displacement
Therefore,
$textbf{v}=textbf{r}'=textbf{i}+2textbf{j}-2ttextbf{j}+3textbf{k}-2ttextbf{k}$ = velocity
And,
$textbf{a}=textbf{r}''=-2textbf{j}-2textbf{k}$ = acceleration
To show orthogonality, I know $textbf{a} cdot textbf{b}=0$, so for velocity and acceleration to be orthogonal (or perpendicular), $textbf{v} cdot textbf{a}=0$.
$textbf{v} cdot textbf{a}= (textbf{i}+2textbf{j}-2ttextbf{j}+3textbf{k}-2ttextbf{k}) cdot (-2textbf{j}-2textbf{k})=[textbf{i}+(2+2t)textbf{j}+(3-2t)textbf{k}] cdot (-2textbf{j}-2textbf{k})=(1)(0)textbf{i}+(2+2t)(-2)textbf{j}+(3-2t)(-2)textbf{k}=-4textbf{j}-4ttextbf{j}-6textbf{k}+4ttextbf{k}=0$
$implies 4ttextbf{j}-4ttextbf{k}=-4textbf{j}-6textbf{k}$
Obviously, there is no value for $t$ which satisfies the above, so I think I may have made a mistake!
Thanks.
vectors orthogonality classical-mechanics
asked May 2 '17 at 6:32
Kyle JohnsonKyle Johnson
275
$endgroup$
So, for $textbf{r}=textbf{r}(t)$,
$textbf{r}=(t+3)textbf{i}+(2t-t^{2})textbf{j}+(3t-t^{2})textbf{k}$ = displacement
Therefore,
$textbf{v}=textbf{r}'=textbf{i}+2textbf{j}-2ttextbf{j}+3textbf{k}-2ttextbf{k}$ = velocity
And,
$textbf{a}=textbf{r}''=-2textbf{j}-2textbf{k}$ = acceleration
To show orthogonality, I know $textbf{a} cdot textbf{b}=0$, so for velocity and acceleration to be orthogonal (or perpendicular), $textbf{v} cdot textbf{a}=0$.
$textbf{v} cdot textbf{a}= (textbf{i}+2textbf{j}-2ttextbf{j}+3textbf{k}-2ttextbf{k}) cdot (-2textbf{j}-2textbf{k})=[textbf{i}+(2+2t)textbf{j}+(3-2t)textbf{k}] cdot (-2textbf{j}-2textbf{k})=(1)(0)textbf{i}+(2+2t)(-2)textbf{j}+(3-2t)(-2)textbf{k}=-4textbf{j}-4ttextbf{j}-6textbf{k}+4ttextbf{k}=0$
$implies 4ttextbf{j}-4ttextbf{k}=-4textbf{j}-6textbf{k}$
Obviously, there is no value for $t$ which satisfies the above, so I think I may have made a mistake!
Thanks.
vectors orthogonality classical-mechanics
vectors orthogonality classical-mechanics
asked May 2 '17 at 6:32
Kyle JohnsonKyle Johnson
275
asked May 2 '17 at 6:32
Kyle JohnsonKyle Johnson
275
asked May 2 '17 at 6:32
Kyle JohnsonKyle Johnson
275
asked May 2 '17 at 6:32
Kyle JohnsonKyle Johnson
275
asked May 2 '17 at 6:32
Kyle JohnsonKyle Johnson
275
275
1
$begingroup$
The dot product if often called the "scalar product". That reminds its users that its output is a scalar not a vector.
$endgroup$
– Lord Shark the Unknown
May 2 '17 at 6:34
add a comment |
1
$begingroup$
The dot product if often called the "scalar product". That reminds its users that its output is a scalar not a vector.
$endgroup$
– Lord Shark the Unknown
May 2 '17 at 6:34
1
1
$begingroup$
The dot product if often called the "scalar product". That reminds its users that its output is a scalar not a vector.
$endgroup$
– Lord Shark the Unknown
May 2 '17 at 6:34
$begingroup$
The dot product if often called the "scalar product". That reminds its users that its output is a scalar not a vector.
$endgroup$
– Lord Shark the Unknown
May 2 '17 at 6:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As you wrote, $textbf v=textbf i+(2-2t)cdot textbf j + (3-2t)cdot textbf k$ and $textbf a=-2cdot textbf j-2cdot textbf k$.
The vector product and the solution when its zero is derived by $$textbf vcdot textbf a =-2(2-2t)-2(3-2t)=0implies t=frac{5}{4}$$
In your solution, you should note that $textbf jcdot textbf j = textbf kcdot textbf k = 1$, and you also miswrote $"(2+2t)(-2)"$ in your equation, which should be $(2-2t)(-2)$.
answered May 2 '17 at 6:36
Lazy LeeLazy Lee
3,4711517
$endgroup$
$begingroup$
Oh I see. Thank you very much for your help!
$endgroup$
– Kyle Johnson
May 2 '17 at 6:50
$begingroup$
Glad to have helped! :) :)
$endgroup$
– Lazy Lee
May 2 '17 at 6:51
add a comment |
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1 Answer
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$begingroup$
As you wrote, $textbf v=textbf i+(2-2t)cdot textbf j + (3-2t)cdot textbf k$ and $textbf a=-2cdot textbf j-2cdot textbf k$.
The vector product and the solution when its zero is derived by $$textbf vcdot textbf a =-2(2-2t)-2(3-2t)=0implies t=frac{5}{4}$$
In your solution, you should note that $textbf jcdot textbf j = textbf kcdot textbf k = 1$, and you also miswrote $"(2+2t)(-2)"$ in your equation, which should be $(2-2t)(-2)$.
answered May 2 '17 at 6:36
Lazy LeeLazy Lee
3,4711517
$endgroup$
$begingroup$
Oh I see. Thank you very much for your help!
$endgroup$
– Kyle Johnson
May 2 '17 at 6:50
$begingroup$
Glad to have helped! :) :)
$endgroup$
– Lazy Lee
May 2 '17 at 6:51
add a comment |
$begingroup$
As you wrote, $textbf v=textbf i+(2-2t)cdot textbf j + (3-2t)cdot textbf k$ and $textbf a=-2cdot textbf j-2cdot textbf k$.
The vector product and the solution when its zero is derived by $$textbf vcdot textbf a =-2(2-2t)-2(3-2t)=0implies t=frac{5}{4}$$
In your solution, you should note that $textbf jcdot textbf j = textbf kcdot textbf k = 1$, and you also miswrote $"(2+2t)(-2)"$ in your equation, which should be $(2-2t)(-2)$.
answered May 2 '17 at 6:36
Lazy LeeLazy Lee
3,4711517
$endgroup$
$begingroup$
Oh I see. Thank you very much for your help!
$endgroup$
– Kyle Johnson
May 2 '17 at 6:50
$begingroup$
Glad to have helped! :) :)
$endgroup$
– Lazy Lee
May 2 '17 at 6:51
add a comment |
$begingroup$
As you wrote, $textbf v=textbf i+(2-2t)cdot textbf j + (3-2t)cdot textbf k$ and $textbf a=-2cdot textbf j-2cdot textbf k$.
The vector product and the solution when its zero is derived by $$textbf vcdot textbf a =-2(2-2t)-2(3-2t)=0implies t=frac{5}{4}$$
In your solution, you should note that $textbf jcdot textbf j = textbf kcdot textbf k = 1$, and you also miswrote $"(2+2t)(-2)"$ in your equation, which should be $(2-2t)(-2)$.
answered May 2 '17 at 6:36
Lazy LeeLazy Lee
3,4711517
$endgroup$
As you wrote, $textbf v=textbf i+(2-2t)cdot textbf j + (3-2t)cdot textbf k$ and $textbf a=-2cdot textbf j-2cdot textbf k$.
The vector product and the solution when its zero is derived by $$textbf vcdot textbf a =-2(2-2t)-2(3-2t)=0implies t=frac{5}{4}$$
In your solution, you should note that $textbf jcdot textbf j = textbf kcdot textbf k = 1$, and you also miswrote $"(2+2t)(-2)"$ in your equation, which should be $(2-2t)(-2)$.
answered May 2 '17 at 6:36
Lazy LeeLazy Lee
3,4711517
answered May 2 '17 at 6:36
Lazy LeeLazy Lee
3,4711517
answered May 2 '17 at 6:36
Lazy LeeLazy Lee
3,4711517
answered May 2 '17 at 6:36
Lazy LeeLazy Lee
3,4711517
3,4711517
$begingroup$
Oh I see. Thank you very much for your help!
$endgroup$
– Kyle Johnson
May 2 '17 at 6:50
$begingroup$
Glad to have helped! :) :)
$endgroup$
– Lazy Lee
May 2 '17 at 6:51
add a comment |
$begingroup$
Oh I see. Thank you very much for your help!
$endgroup$
– Kyle Johnson
May 2 '17 at 6:50
$begingroup$
Glad to have helped! :) :)
$endgroup$
– Lazy Lee
May 2 '17 at 6:51
$begingroup$
Oh I see. Thank you very much for your help!
$endgroup$
– Kyle Johnson
May 2 '17 at 6:50
$begingroup$
Oh I see. Thank you very much for your help!
$endgroup$
– Kyle Johnson
May 2 '17 at 6:50
$begingroup$
Glad to have helped! :) :)
$endgroup$
– Lazy Lee
May 2 '17 at 6:51
$begingroup$
Glad to have helped! :) :)
$endgroup$
– Lazy Lee
May 2 '17 at 6:51
add a comment |
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1
$begingroup$
The dot product if often called the "scalar product". That reminds its users that its output is a scalar not a vector.
$endgroup$
– Lord Shark the Unknown
May 2 '17 at 6:34