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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Mar 2008 10:53:39 IST
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A particle moves such that its position vector r at time t is given by: r=(3t2+2t)i?+(2t2+4t)?j ; determine the magnitude and direction of its velocity and acceleration at t = 5 sec.
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29 Mar 2008 11:07:18 IST
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Velocity vector = dr/dt = (9t2+2)i + (4t+4)j
at t=5, v = (227i + 40j).
Magnitude =  2272+402
Direction = tan-1(40/227) with the positive direction of x axis)
Similarly, calculate acceleration = d2r/dt2
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Will nip in at times to solve problems :)
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30 Mar 2008 08:26:15 IST
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Excuse me, Karthik,
Could you please explain, how th derivative of (3t2 + 2t)i is equal to (9t2+2)i.
It should be equal to (6t+2)i
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30 Mar 2008 09:49:09 IST
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r = (3t2 + 2t) i + (2t2 + 4t) j
differentiating wrt. to time we get
v= (6t + 2) i + (4t+ 4 ) j ----------- 1
putting t= 5 we get
v= 32i + 24j
taking magnitude = underroot of ( 32^2 +24^2 ) = 40...
differentiating eq. 1 wrt. to time we get
a = 6i + 4j
taking magnitude we get = under root of (6^2+ 4^2 ) = 7.41...
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IMPOSSIBLES ARE OFTEN UNTRIED... |
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30 Mar 2008 21:07:27 IST
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Excuse me Akshay,
The answers which i got were ,
v = 40 m/s , Theta = 36.86 w.r.t x - axis, &
a = 7.2 m/s2 , Theta = 33.36 w.r.t x - axis.
Could you please verify the direction part.
Thank You
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