plz explain in detail
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a = dv/dt = 2a -6bt
2a -6bt = 0
t = 2a/6b = a/3b...
Given : x = at² - bt³
Acceleration is change in velocity per unit time [ a = dv/dt ].
also,
Velocity is change in displacement per unit time [ v = dx/dt ].
Thus,
Acceleration = d²x/dt²
now,
differentiate x = at² - bt³ with respect to t.
dx/dt = 2at - 3bt²
again,
d²x/dt² = 2a - 6bt
acc. to above we know, A. = d²x/dt²
Acceleration = 2a - 6bt
but, ATQ
2a - 6bt = 0
6bt = 2a
t = a/3b
Hence, Time ( t ) = a / 3b
Option [ 3 ] : a/3b is CORRECT ! ! !
Hope it helps! :D