a particle starts fr rest and acceleration at any time is given by f-kt² where f and t are constants. calculate the maximum velocity u of the particle during motion and distance travelled by it before it acquires this velocity
Answers
maximum velocity = √(f/k)(2f/3) and distance covered till then = 3f²/7k
Explanation:
a = f - kt²
a = dv/dt
=> dv/dt = f - kt²
=> v = ft - kt³/3 + c
at t= 0 v = o as start from rest
Hene c = 0
=> v = ft - kt³/3
a = dv/dt = f - kt²
da/dt = d²v/dt² = - 2kt < 0
hence velocity is maximum when dv/dt = 0
=> f - kt² = 0
=> t² = f/k
=> t = √(f/k)
v = ft - kt³/3
v max at t = √(f/k)
=> v = t ( f - kt²/3)
=> v = √(f/k) ( f - f/3)
=> v = √(f/k)(2f/3)
v = ft - kt³/3
v = ds/dt
ds = ( ft - kt³/3 ) dt
boundary t = 0 & t = √(f/k)
s = ft²/2 - kt⁴/14
s = t²/2 (f - kt²/7)
s = (f/2k)( f - f/7)
=> s = (f/2k)( 6f/7)
=> s = 3f²/7k
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