Asked by borjabbol
                A particle is moving along a line according to the equation of motion s = 0.5 t^2 + 4t / (t + 1), where s is in meters and t is in seconds. What is the instantaneous velocity when the acceleration is zero?
            
            
        Answers
                    Answered by
            mathhelper
            
    s = 0.5 t^2 + 4t / (t + 1)
v = ds/dt = t + ( (t+1)(4) - 4t(1) )/(t+1)^2
= t + 4/(t+1)^2
a = dv/dt = 1 -8(t+1)^-3 = 1 - 8/(t+1)^3
= 0
1 = 8/(t+1)^3
(t+1)^3 = 8
t + 1 = 2
t = 1 s
then when t = 1,
v = 1 + 4/2^2 = 2 m/s
    
v = ds/dt = t + ( (t+1)(4) - 4t(1) )/(t+1)^2
= t + 4/(t+1)^2
a = dv/dt = 1 -8(t+1)^-3 = 1 - 8/(t+1)^3
= 0
1 = 8/(t+1)^3
(t+1)^3 = 8
t + 1 = 2
t = 1 s
then when t = 1,
v = 1 + 4/2^2 = 2 m/s
                    Answered by
            oobleck
            
    v = ds/dt = t + 4/(t+1)^2
a = 1 - 8/(t+1)^3
a=0 when t=1
v(1) = 2
    
a = 1 - 8/(t+1)^3
a=0 when t=1
v(1) = 2
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