Question

the displacement x(t) of a particle from the origin at time t, t≥0, is x(t)=t^3+3t(t-3)+10. Find min velocity?

Answer 1

Answer:

0 m/s

Explanation:

given,

x(t)=t^3+3 t (t-3)+10

velocity

$v = \dfrac{dx}{dt}$

$\dfrac{dx}{dt} =\dfrac{d}{dt}(t^3+3 t (t-3)+10)$

$v=3t^2 +3t + (t-3)3$.............(1)

now, differentiating both side for minimum velocity calculation

$\dfrac{dv}{dt}=\dfrac{d}{dt}(3t^2 +3t + (t-3)3)$

$\dfrac{dv}{dt}=6t + 6$

now, equating derivative equal to zero

$\dfrac{dv}{dt}=0$

$6t + 6=0$

t = -1

again differentiating

$\dfrac{d^2v}{dt^2}=+ve$

hence, at t = -1 velocity will be minimum

now, inserting value of t in equation 1

$v=3(1)^2 +3\times 1 + (1-3)3$

$v=6-6$

 v = 0 m/s

hence, minimum velocity is equal to 0 m/s