Angle between the positive direction of the x axis and a line tangent to the particle's path
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The position $vec r$ of a particle moving in an xy plane is given by $vec r$ = $(1.00t^3 − 3.00t)i + (8.00 − 9.00t^4)j$, with $vec r$ in meters and $t$ in seconds. In unit-vector notation, calculate the following for $t = 2.10 s$.
What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 2.10 s? (counterclockwise from the +x-axis)
Is it asking for a degree based on r, v, or a?
Since it mentioned s, I thought it referred to v. So, I did:
-333.4/10.28 times tan inverse. That got -88.1, then I added 180 degrees since it was in Q4, for 91.76 degrees. It says it's wrong, though.
physics
asked Sep 24 '16 at 4:05
user366783user366783
3829
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add a comment |
$begingroup$
The position $vec r$ of a particle moving in an xy plane is given by $vec r$ = $(1.00t^3 − 3.00t)i + (8.00 − 9.00t^4)j$, with $vec r$ in meters and $t$ in seconds. In unit-vector notation, calculate the following for $t = 2.10 s$.
What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 2.10 s? (counterclockwise from the +x-axis)
Is it asking for a degree based on r, v, or a?
Since it mentioned s, I thought it referred to v. So, I did:
-333.4/10.28 times tan inverse. That got -88.1, then I added 180 degrees since it was in Q4, for 91.76 degrees. It says it's wrong, though.
physics
asked Sep 24 '16 at 4:05
user366783user366783
3829
$endgroup$
add a comment |
$begingroup$
The position $vec r$ of a particle moving in an xy plane is given by $vec r$ = $(1.00t^3 − 3.00t)i + (8.00 − 9.00t^4)j$, with $vec r$ in meters and $t$ in seconds. In unit-vector notation, calculate the following for $t = 2.10 s$.
What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 2.10 s? (counterclockwise from the +x-axis)
Is it asking for a degree based on r, v, or a?
Since it mentioned s, I thought it referred to v. So, I did:
-333.4/10.28 times tan inverse. That got -88.1, then I added 180 degrees since it was in Q4, for 91.76 degrees. It says it's wrong, though.
physics
asked Sep 24 '16 at 4:05
user366783user366783
3829
$endgroup$
The position $vec r$ of a particle moving in an xy plane is given by $vec r$ = $(1.00t^3 − 3.00t)i + (8.00 − 9.00t^4)j$, with $vec r$ in meters and $t$ in seconds. In unit-vector notation, calculate the following for $t = 2.10 s$.
What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 2.10 s? (counterclockwise from the +x-axis)
Is it asking for a degree based on r, v, or a?
Since it mentioned s, I thought it referred to v. So, I did:
-333.4/10.28 times tan inverse. That got -88.1, then I added 180 degrees since it was in Q4, for 91.76 degrees. It says it's wrong, though.
physics
physics
asked Sep 24 '16 at 4:05
user366783user366783
3829
asked Sep 24 '16 at 4:05
user366783user366783
3829
asked Sep 24 '16 at 4:05
user366783user366783
3829
asked Sep 24 '16 at 4:05
user366783user366783
3829
asked Sep 24 '16 at 4:05
user366783user366783
3829
3829
add a comment |
add a comment |
1 Answer
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"Direction" refers to velocity. So calculate the derivative vector at time $t=2.1$ seconds.
answered Sep 24 '16 at 4:10
bubbabubba
30.7k33188
$endgroup$
$begingroup$
So I should do -333.3/10.28? I already tried that.
$endgroup$
– user366783
Sep 24 '16 at 4:38
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
"Direction" refers to velocity. So calculate the derivative vector at time $t=2.1$ seconds.
answered Sep 24 '16 at 4:10
bubbabubba
30.7k33188
$endgroup$
$begingroup$
So I should do -333.3/10.28? I already tried that.
$endgroup$
– user366783
Sep 24 '16 at 4:38
add a comment |
$begingroup$
"Direction" refers to velocity. So calculate the derivative vector at time $t=2.1$ seconds.
answered Sep 24 '16 at 4:10
bubbabubba
30.7k33188
$endgroup$
$begingroup$
So I should do -333.3/10.28? I already tried that.
$endgroup$
– user366783
Sep 24 '16 at 4:38
add a comment |
$begingroup$
"Direction" refers to velocity. So calculate the derivative vector at time $t=2.1$ seconds.
answered Sep 24 '16 at 4:10
bubbabubba
30.7k33188
$endgroup$
"Direction" refers to velocity. So calculate the derivative vector at time $t=2.1$ seconds.
answered Sep 24 '16 at 4:10
bubbabubba
30.7k33188
answered Sep 24 '16 at 4:10
bubbabubba
30.7k33188
answered Sep 24 '16 at 4:10
bubbabubba
30.7k33188
answered Sep 24 '16 at 4:10
bubbabubba
30.7k33188
30.7k33188
$begingroup$
So I should do -333.3/10.28? I already tried that.
$endgroup$
– user366783
Sep 24 '16 at 4:38
add a comment |
$begingroup$
So I should do -333.3/10.28? I already tried that.
$endgroup$
– user366783
Sep 24 '16 at 4:38
$begingroup$
So I should do -333.3/10.28? I already tried that.
$endgroup$
– user366783
Sep 24 '16 at 4:38
$begingroup$
So I should do -333.3/10.28? I already tried that.
$endgroup$
– user366783
Sep 24 '16 at 4:38
add a comment |
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