A point travels along the x axis with a velocity whose projection `v_x` is presented as a function of time by the plot in figure.

Assuming the coordinate of the point `x=0` at the moment `t=0`, draw the approximate time dependence plots for the acceleration `w_x`, the x coordinate, and the distance covered s.

Answer

`(##IROD_V01_C01_E01_018_S01##)`

Explanation

To plot `x(t), s(t)` and `w_x(t)` let us partion the given plot `v_x(t)` into five segments (for detailed analysis) as shown in the figure.
For the part `oa:w_x=1` and `v_x=t=v`
Thus, `Deltax_1(t)=int v_xdt-underset0oversettintdt-t^2/2-s_1(t)`
Putting `t=1`, we get, `Deltax_1=s=1/2unit`
For the part ab:
`w_x=O` and `nu_x=nu=const ant=1`
Thus `Deltax_2(t)=intv_xdt=underset1oversettintdt=(t-1)=s_2(t)`
Putting `t=3`, `Deltax_2=s_2=2units`
For the part `b4: w_x=1` and `v_x=1-(t-3)=4-t)=v`
Thus `Deltax_3(t)=underset3oversettint(4-t)dt=4t-t^2/2-(15)/(2)=s_3(t)`
Putting `t=4, Deltax_3=x_3=1/2unit`
For the part `4d: v_x=-1` and `v_x=-(1-4)=4-1`
So, `v=|v_x|=t-4` for `tgt4`
Thus `Deltax_4(t)=underset4oversettint(1-t)dt=4t-t^2/2-8`
Putting `t=6, Deltax_4=-1`
Similarly `s_4(t)=int |v_x|dt=underset4oversettint(t-4)dt=t^2/2-4t+8`
Putting `t=6, s_4=2unit`
For the part `d7: w_x=2` and `v_x=-2+2(t-6)=2(t-7)`
`v=|v_x|=2(7-t)` for `tlarr7`
Now, `Deltax(t) = undersett overset6int2(t-7)dt=t^2-14t+48`
Putting `t=4`, `Deltax_5=-1`
Similarly `s_5(t)=undersett overset6int2(7-t)dt=14t-t^2-48`
Putting `t=7, s_5=1`
On the basis of these obtained expressions `w_x(t), x(t)` and `s(t)` plots can be easily plotted as shown in the figure of answersheet.