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pLuvia

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4. A particle is in oscillating motion obeying the law x=5cos2t. Find the velocity and acceleration at the points x=-5 and x=0

8. A particle passes a fixed point O and moves in a stragiht line with a velocity of (8+[1/3] t2) m/s, where t is the time in seconds after passing O. Find the distance travlled during the fourth second
 

Riviet

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pLuvia said:
4. A particle is in oscillating motion obeying the law x=5cos2t. Find the velocity and acceleration at the points x=-5 and x=0
Find dx/dt and substitute x=-5 and 0 to find velocity at each of those points, then differentiate dx/dt and sub x=-5 and 0 again to find acceleration at each of these points.
 

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pLuvia said:
8. A particle passes a fixed point O and moves in a stragiht line with a velocity of (8+[1/3] t2) m/s, where t is the time in seconds after passing O. Find the distance travlled during the fourth second
Oo, i just figured out the second too lol. Just integrate 8+[1/3] t2 and sub in t=4 to find the distance travelled.
 
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pLuvia

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But if I differentiate x=5cos2t there's no x after that :confused:
 
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pLuvia

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Yeh I did that way as well, but that's not the answer :(

I did that and got 352/9 but the answer is 109/9 ><"
 

Templar

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pLuvia said:
But if I differentiate x=5cos2t there's no x after that :confused:
You differentiate with respect to t.

pLuvia said:
I did that and got 352/9 but the answer is 109/9 ><"
Travelled in the fourth second means the distance travelled from t=3 to t=4, so do a definite integral using those values.
 

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Templar said:
You differentiate with respect to t.
Exactly.

So you should get dx/dt = -10sin2t, and then just sub in the x values.
 

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