Respuesta :
First compute the resultant force F:
[tex]\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N[/tex]
[tex]\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N[/tex]
[tex]\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N[/tex]
Then use Newton's second law to determine the acceleration vector [tex]\mathbf a[/tex] for the particle:
[tex]\mathbf F=m\mathbf a[/tex]
[tex](10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a[/tex]
[tex]\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}[/tex]
Let [tex]\mathbf x(t)[/tex] and [tex]\mathbf v(t)[/tex] denote the particle's position and velocity vectors, respectively.
(a) Use the fundamental theorem of calculus. The particle starts at rest, so [tex]\mathbf v(0)=0[/tex]. Then the particle's velocity vector at t = 10.4 s is
[tex]\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du[/tex]
[tex]\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}[/tex]
[tex]\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}[/tex]
If you don't know calculus, then just use the formula,
[tex]v_f=v_i+at[/tex]
So, for instance, the velocity vector at t = 10.4 s has x-component
[tex]v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}[/tex]
(b) Compute the angle [tex]\theta[/tex] for [tex]\mathbf v(10.4\,\mathrm s)[/tex]:
[tex]\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ[/tex]
so that the particle is moving at an angle of about 313º counterclockwise from the positive x axis.
(c) We can find the velocity at any time t by generalizing the integral in part (a):
[tex]\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du[/tex]
[tex]\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j[/tex]
Then using the fundamental theorem of calculus again, we have
[tex]\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du[/tex]
where [tex]\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m[/tex] is the particle's initial position. So we get
[tex]\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du[/tex]
[tex]\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}[/tex]
[tex]\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m[/tex]
So over the first 10.4 s, the particle is displaced by the vector
[tex]\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m[/tex]
or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).
(d) See part (c).
