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Sagot :
We know that the acceleration is defined as:
[tex]a=\frac{dv}{dt}[/tex]This means that if we know the acceleration we can find the velocity with the integral:
[tex]v=\int ^t_{t_0}adt^{\prime}+v_0[/tex]where t0 denotes some intial time and v0 the velocity at that time.
In this case we know that the acceleration is 4m/s^2 and that at time t=1 the velocity is 4 m/s, then we have:
[tex]\begin{gathered} v=\int ^t_14dt^{\prime}+4 \\ =4(t^{\prime})\vert^t_1+4 \\ =4(t-1)+4 \\ =4t-4+4 \\ =4t \end{gathered}[/tex]hence the velocity function for this motion is:
[tex]v=4t[/tex]Now, we know that the velocity is defined by:
[tex]v=\frac{dx}{dt}[/tex]then the position can be obtained by:
[tex]x=\int ^t_{t_0}vdt^{\prime}+x_0[/tex]where t0 is some time and x0 is the position at that time; in this case we don't have an initial position for the particle then te position will be given by:
[tex]\begin{gathered} x=\int ^t_{t_0}4t^{\prime}dt^{\prime}+x_0 \\ x=4(\frac{t^{\prime2}}{2})\vert^t_{t0}+x_0 \\ x=4(\frac{t^2}{2}-\frac{t^2_0}{2})+x_0 \\ x=2t^2-2t^2_0+x_0 \end{gathered}[/tex]Hence the position at any given time is given by:
[tex]x=2t^2-2t^2_0+x_0[/tex]Once we know the position function we can calculate the area undert the position time graph in the interval given:
[tex]\begin{gathered} \int ^3_1(2t^2-2t^2_0+x_0)dt=(\frac{2}{3}t^3+(x_0-2t^2_0)t)\vert^3_1 \\ =\frac{2}{3}(3^3-1^3)+(x_0-2t^2_0)(3^{}-1^{}) \\ =\frac{2}{3}(27-1)+(x_0-2t^2_0)(2) \\ =\frac{2}{3}(26)+(x_0-2t^2_0)(2) \\ =\frac{52}{3}+2(x_0-2t^2_0) \end{gathered}[/tex]Therefore, in general, the area under the curve in the given interval is:
[tex]\frac{52}{3}+2(x_0-2t^2_0)[/tex]If we assume that the particle was at the origin at time t0=0 (this meas x0=0 as well), then the area under the position-time graph will be:
[tex]\frac{52}{3}+2(0-2(0)^2)=\frac{52}{3}[/tex]Note: No matter what the intial time and position is the expression we found will give us the correct answer.
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