Looking for classes? Ksquare Career Institute, Bengaluru →
A particle moves along a straight line OX. At a time $t$ (in seconds) the distance $x$ (in metres) of the particle from O is given by $x = 40 + 12t - t^3$. How long would the particle travel before coming to rest :-
A
24 m
B
40 m
C
56 m
D
16 m
Detailed Solution
$x = 40 + 12t - t^3$
Velocity: $v = \dfrac{dx}{dt} = 12 - 3t^2$
The particle comes to rest when $v = 0$: $12 - 3t^2 = 0 \Rightarrow t^2 = 4 \Rightarrow t = 2$ s
Distance travelled by the particle before coming to rest $= x(\text{at } t = 2) - x(\text{at } t = 0)$
$x(2) = 40 + 12 \times 2 - 2^3 = 40 + 24 - 8 = 56$ m
$x(0) = 40$ m
Distance travelled $= 56 - 40 = 16$ m
Velocity: $v = \dfrac{dx}{dt} = 12 - 3t^2$
The particle comes to rest when $v = 0$: $12 - 3t^2 = 0 \Rightarrow t^2 = 4 \Rightarrow t = 2$ s
Distance travelled by the particle before coming to rest $= x(\text{at } t = 2) - x(\text{at } t = 0)$
$x(2) = 40 + 12 \times 2 - 2^3 = 40 + 24 - 8 = 56$ m
$x(0) = 40$ m
Distance travelled $= 56 - 40 = 16$ m