A particle moves in a straight line and at point P, it's velocity is given as v = 7t^2 - 5t +3. The particle comes to rest at point Q.
1. What is the acceleration at Q if it arrives at Q when t=7?
2. How far does the particle travel in t=1 to t=4?
Sunday, April 19, 2009
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well i can work the acceleration of the particle
ReplyDeletewhere v = 7t^2 -5t +3
dv/dt= 14t - 5
if the particle comes to rest at point Q, then there shouldn't b any acceleration at that point.
therefore acceleration should be = 0
ok i really don't know.....can some1 help me
I think your on the right track with the acceleration part But am also lost with the acceleration at point Q.
ReplyDeleteDo you think that it suppose to be dv/dQ we suppose to be finding, using a substitution....
Am not sure here guys, but help me out....
For part 2, i know that when you find dv/dt all you have to do now is to substitute the value of t=1 and t=4 in it to find how far the perticle travelled..
ReplyDeletei just think it is a trick question becasue at rest the acceleration will be 0 so i agree with crazykid
ReplyDeletePost corrected.
ReplyDeleteSwanky Thing, u'r wrong.To find dist. you need to intergrate velocity..
ReplyDeleteso: v= 7t^2 -5t +3
s= 7t^3/3 - 5t^2/2 + 3t