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A particle started moving in a straight line.Its acceleration at time t seconds is given by a=(-5t^(2)+5)m/s^(2) t>=0 .Find the time for its maximum velocity and maximum velocity of the particle v(max)?

Asked by sharadkumar12055 24th May 2021, 4:07 PM
Answered by Expert
Answer:
acceleration a = begin mathsize 14px style fraction numerator d v over denominator d t end fraction end style = -5t2 + 5 ...................(1)
maximum time t is determined by equating eqn.(1) to zero and solve for time t
 
-5t2 + 5 = 0
 
from above expression , we get , tmax = 1 s
 
Velocity as a function of time t is determined by integrating eqn.(1)
 
begin mathsize 14px style v left parenthesis t right parenthesis space equals space integral subscript 0 superscript t d v space equals space minus 5 integral subscript 0 superscript t tau squared d tau space plus space 5 integral subscript 0 superscript t d tau space end style
v(t) = -(5/3) t3 + 5 t
 
vmax is determined by substituting t =1 in above equation
 
vmax = -(5/3) + 5 = ( 10/3 ) m/s
Answered by Expert 24th May 2021, 5:04 PM
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