Complete Question
The complete question is shown on the first uploaded image
Answer:
The particle's position is [tex]x_{2} = 10 m[/tex]
The particle's velocity is [tex]v = 2 \ m/s[/tex]
Explanation:
From the question we are told that
[tex]x = 2m[/tex] at [tex]t_o = 0 \ sec[/tex]
and from the graph at t = 0 [tex]v = 6 /m[/tex]
Now the acceleration which is the slope of the graph is mathematically represented as
[tex]a = - \frac{6 - 4}{3-2}[/tex]
[tex]a = - 2 m/s^2[/tex]
The negative sign shows that it is a negative slope
Now to obtain the velocity at t = 2 sec
We use the equation of motion as follows
[tex]v = v_o + at[/tex]
substituting values '
[tex]v = 6 + (-2)(2)[/tex]
[tex]v = 2 \ m/s[/tex]
Now to obtain the position of the particle at v = 2 m/s
We use the equation of motion as follows
[tex]v^2 = v_o ^2 + 2 ax[/tex]
So [tex]2 ^2 = 6^2 + 2(-2)x[/tex]
[tex]4x = 32[/tex]
[tex]x = 8 m[/tex]
From above [tex]x = 2m[/tex] at [tex]t_o = 0 \ sec[/tex]
So the position at t = 2 s
[tex]x_{2} = x + x_o[/tex]
[tex]x_{2} = 2 + 8[/tex]
[tex]x_{2} = 10 m[/tex]
