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 = 
= -5t2 + 5 ...................(1)
= -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)
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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