SOLUTION: A particle moves in a straight line so that its distance, s m, from a fixed point A on the line is given by s=2t^2-4t+9, for t less than or equal to 3, where t is the time in secon

Algebra ->  Test -> SOLUTION: A particle moves in a straight line so that its distance, s m, from a fixed point A on the line is given by s=2t^2-4t+9, for t less than or equal to 3, where t is the time in secon      Log On


   

Question 1081862: A particle moves in a straight line so that its distance, s m, from a fixed point A on the line is given by s=2t^2-4t+9, for t less than or equal to 3, where t is the time in seconds after passing through a point B on the line.
(a) the total distance travelled by the particle in the period of t=0 to t=3. Ans=10m
(b) At t=3, the acceleration of the particle is changed to (t − 8)m/s2 , the instantaneous velocity remaining unchanged. Find the next value of t at which the particle comes to instantaneous rest. Ans= 5s
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
a)
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Looking at the graph, the particle travels from 9 to 7 in the 1st second, from 7 to 9 in the 2nd second and from 9 to 15 in the 3rd second.
So it traveled,
D=abs%287-9%29%2Babs%289-7%29%2Babs%2815-9%29=2%2B2%2B6=10m
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b) Find the instantaneous velocity when t=3,
v=ds%2Fdt=4t-4=4%283%29-4=12-4=8m%2Fs
So then integrate the new acceleration to get the new velocity,
v=int%28a%2Cdt%2Ct=3%2Ct%29=int%28%28t-8%29%2Cdt%2Ct=3%2Ct%29=t%5E2%2F2-8t%2BC
When t=3,v=8,
8=%283%29%5E2%2F2-8%283%29%2BC
8=9%2F2-24%2BC
C=32-9%2F2
C=64%2F2-9%2F2
C=55%2F2
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v%28t%29=t%5E2%2F2-8t%2B55%2F2 for t%3E=3
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Graphing this and looking for the next crossing of the x-axis,
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t=5s
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You could also factor,
t%5E2%2F2-8t%2B55%2F2=0
t%5E2-16t%2B55=0
%28t-5%29%28t-11%29=0
t=5 and t=11