A particle moves along a straight line $OX$. At a time $t$ (in second) the distance $x$ (in metre) of the particle from $O$ is given by
$ x=40+12t-t^{3} $
How long would the particle travel before coming to rest?
- 24 m
- 40 m
- 56 m
- 16 m
The Correct Option is C
Solution and Explanation
$x=40+12 t-t^{3}$
We know that, speed is rate of change of distance
$i e, v=\frac{d x}{d t}$
$\therefore v =\frac{d}{d t}\left(40+12 t-t^{3}\right) $
$=0+12-3 t^{2}$
but final velocity $v=0$
$\therefore 12-3 t^{2}=0 $
or $ t^{2}=\frac{12}{3}=4$
or $ t=2 \,s$
Hence, distance travelled by the particle before coming to rest is given by
$x =40+12(2)-(2)^{3} $
$=40+24-8=64-8 $
$=56 \,m$
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Concepts Used:
Distance and Displacement
Distance:
The sum of the length of the path traveled by an object from one place to another is called distance. The path may or may not be directly from the initial point to the final point.
Distance is a scalar quantity and has only magnitude, also does not have any direction.
For example,
From the particular point, if a car travels to the east for 5 km and takes a turn to travel north for another 8 km, the total distance traveled by car shall be 13 km. The distance can never be zero or negative but should be always more than the displacement of the object. The distance of the object gives complete information about the path traveled by the object.
Read More: Difference between Distance and Displacement
Displacement:
The length of the shortest path from the initial point to the final point is called displacement. It is a vector quantity that consists of magnitude as well as direction.
For example,
Let's consider the same example given above, the total displacement of the object will be the length of the line joining the two positions. The displacement of an object is usually shorter or equal to the distance traveled by the object. The displacement of the object does not give the proper information about the path traveled by the object.