A particle moves along a straight line OX. At a time t (in second) the distance x of the particle from O is given by x=40+12t-t3. How long would the particle travel before coming to rest?

Question

A particle moves along a straight line OX. At a time t (in second) the distance x of the particle from O is given by x=40+12t-t3. How long would the particle travel before coming to rest? (A) 24 m (B) 40 m (C) 56 m (D) 16 m

Tardigrade Admin · Accepted Answer

Distance travelled by the particle is x=40+12t-t3 v=(dx/dt)=12-3t2 But final velocity v=0 12-3t2=0 t2=(12/3)=4 t=2S x=40+12(2)-8 =56 m

Q. A particle moves along a straight line OX. At a time t (in second) the distance x of the particle from O is given by $x = 40 + 12 t - t^{3} .$ How long would the particle travel before coming to rest?

24 m

40 m

56 m

16 m

Distance travelled by the particle is
$x = 40 + 12 t - t^{3}$
$v = \frac{d x}{d t} = 12 - 3 t^{2}$
But final velocity $v = 0$
$12 - 3 t^{2} = 0$
$t^{2} = \frac{12}{3} = 4$
$t = 2 S$ $x = 40 + 12 (2) - 8$
$= 56 m$

Q. A particle moves along a straight line OX. At a time t (in second) the distance x of the particle from O is given by x=40+12t−t3. How long would the particle travel before coming to rest?