Question

the position x of a particle varies with time (t) as x=at2-bt3.The acceleration at time t of the particle will be equal to zero, where t is equal t​

Answer 1

Answer:

$\boxed{\mathfrak{t = \dfrac{a}{3b}}}$

Explanation:

Position (x) of particle w.r.t. time (t):

$\rm x = at ^{2} - bt ^{3}$

Rate of change of position of particle w.r.t. time is velocity (v):

$\rm \implies v = \dfrac{dx}{dt} \\ \\ \rm \implies v = \dfrac{d}{dt} (a {t}^{2} - b {t}^{3} ) \\ \\ \rm \implies v = 2at - 3 b{t}^{2}$

Rate of change of velocity of particle w.r.t. time is accelration (a):

$\rm \implies a = \dfrac{dv}{dt} \\ \\ \rm \implies a = \dfrac{d}{dt} (2at - 3b {t}^{2} ) \\ \\ \rm \implies a =2a - 6bt$

Time (t) at which accelration (a) of particle is equal to zero:

$\rm \implies 2a - 6bt = 0 \: \: \: \: \: \: \: \: (a = 0) \\ \\ \rm \implies 6bt = 2a \\ \\ \rm \implies t = \dfrac{2a}{6b} \\ \\ \rm \implies t = \dfrac{a}{3b}$