91Ó°ÊÓ

Initial position and time of the particle is $x_0=0m$and $t_0=0s$.

The turning point is a point in the motion where a particle reverses direction. The velocity instantaneously becomes zero at this point while the position is minimum or maximum.

Here, from the given graph it can be observed that the particle reverses the direction at $t=1s$when $v_x$changes the sign.

Therefore, this particle has a turning point at $t=1s$.

The displacement of a particle $∆x=x_f-x_i$between $t_0$and $t_f$is the area under the velocity-time curve from $t_0$to $t_f$.

Thus, the area under the velocity-time curve is

$∆x=$Area of the triangle between $0s$and $1s$$+$Area of the triangle between $1s$and localid="1648469287677" $2s$

localid="1648469404395" $$\begin{gathered} ∆x=-\frac{1}{2}\left(1s\right)\left(4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)+\frac{1}{2}\left(1s\right)\left(4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right) \\ =0m \end{gathered}$$

Now,

The position of a particle at $t=2s$is given by

$x_f=$$x_0+$area under the velocity curve between role="math" localid="1649476022081" $0s$and role="math" localid="1649476029620" $2s$

$$\begin{gathered} x_f=10m+0m \\ =10m \end{gathered}$$

Therefore, the position of a particle at $t=2s$is $10m$..

The area under the velocity-time curve is

$∆x=$area of the triangle between $0s$and $1s$+ area of the triangle between $1s$and $4s$

localid="1648470524318" $$\begin{gathered} ∆x=-\frac{1}{2}\left(1s\right)\left(4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)+\frac{1}{2}\left(3s\right)\left(12\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right) \\ =-2+18 \\ =16m \end{gathered}$$

Now,

The position of a particle at $t=4s$is given by$x_f=x_0+∆x$

area under the velocity curve between $0s$and$4s$

$$\begin{gathered} x_f=0+16m \\ =16m \end{gathered}$$

Therefore, the position of a particle at $t=4s$is $16m$.