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| For Physics Gurus Only by adebayo18015(op): 5:58am On Oct 27, 2020 |
a particle moves along a curve whose parametric equations are. x=e^-t, y=2cos3t, z=2sin3t {where t is the time } a. determine its velocity and acceleration at any time b. find the magnitude of the velocity and acceleration at t=0 |
| Re: For Physics Gurus Only by NDSMELODY(m): 7:19am On Oct 27, 2020 |
adebayo18015:
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| Re: For Physics Gurus Only by Aluminumquadri(m): 7:22am On Oct 27, 2020 |
x=e^-t, y=2cos3t, z=2sin3t, where x,
y, z are positional vectors
dependent on t.
x'=-e^-t, y'=-6sin3t, z'=6cos3t
where ' denotes d/dt (speed/
velocity).
x"=e^-t, y"=-18cos3t, z"=-18sin3t
(acceleration). When t=0 these are
x"=1, y"=-18, z"=0.
The magnitude of the acceleration
denoted by " is:
√(x"^2+y"^2+z"^2)=√
(1+324)=√325=5√13=18.0278 |
| Re: For Physics Gurus Only by Aluminumquadri(m): 7:24am On Oct 27, 2020 |
x=e^-t, y=2cos3t, z=2sin3t, where x,
y, z are positional vectors
dependent on t.
x'=-e^-t, y'=-6sin3t, z'=6cos3t
where ' denotes d/dt (speed/
velocity).
x"=e^-t, y"=-18cos3t, z"=-18sin3t
(acceleration). When t=0 these are
x"=1, y"=-18, z"=0.
The magnitude of the acceleration
denoted by " is:
√(x"^2+y"^2+z"^2)=√
(1+324)=√325=5√13=18.0278 |
| Re: For Physics Gurus Only by NDSMELODY(m): 7:25am On Oct 27, 2020 |
Last part
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| Re: For Physics Gurus Only by adebayo18015(op): 6:52pm On Oct 30, 2020 |
Aluminumquadri:A very big thank you for this |
| Re: For Physics Gurus Only by adebayo18015(op): 6:52pm On Oct 30, 2020 |
NDSMELODY:Thank you for this |
| Re: For Physics Gurus Only by adebayo18015(op): 6:53pm On Oct 30, 2020 |
[quote author=NDSMELODY post=95381356][/quote]Thanks for this 1 Like |
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