91Ó°ÊÓ

We are given the force in two dimensions in the form

$\stackrel{→}{F}=\left(-2.0t\hat{i}+4.0t^2\hat{j}\right)\mathrm{N}$.

$m$is the mass of the projectile$=2.0kg$

Initial velocity$\stackrel{→}{v}=8.0\widehat{im}/\mathrm{s}$

Let's find the acceleration for component $a_x$and $a_y$:

According to Newton's law, the power exerted on a object creates its accelerates and this force equal to the mass:

Therefore,

$\stackrel{→}{a_x}=\frac{{\stackrel{→}{F}}_x}{m}=\frac{-2.0t\hat{i}}{2.0\mathrm{kg}}=-1.0t\hat{i}\mathrm{m}/{\mathrm{s}}^2$

$\stackrel{→}{a_y}=\frac{{\stackrel{→}{F}}_y}{m}=\frac{4.0t^2\hat{j}}{2.0\mathrm{kg}}=2.0t^2\hat{j}\mathrm{m}/{\mathrm{s}}^2$

Let's compute the final velocity as follow:

$\stackrel{→}{v_f}={\stackrel{→}{v}}_i+{∫}_0^{t}adt$

For the horizontal component:

$=8.0\mathrm{m}/\mathrm{s}\widehat{i}+{∫}_0^{2\mathrm{s}} (−1.0t\widehat{i})dt$

$=8.0\mathrm{m}/\mathrm{s}\widehat{i}+{\left[\frac{−1}{2}{t}^2\right]}_0^{2\mathrm{s}}$

$=8.0\mathrm{m}/\mathrm{s}\widehat{i}+(−2.0\mathrm{m}/\mathrm{s}\widehat{i})$

$=6.0\mathrm{m}/\mathrm{s}\widehat{i}$

For the vertical component:

${\stackrel{→}{v}}_y={\stackrel{→}{v}}_i+{∫}_0^t{a}_{y}dt$

$=0\mathrm{m}/\mathrm{s}\hat{j}+{∫}_0^{2\mathrm{s}}\left(2.0t^2\hat{j}\right)dt$

$={\left[\frac{2}{3}{t}^3\hat{j}\right]}_0^{2\mathrm{s}}$

$=5.33\mathrm{m}/\mathrm{s}\hat{j}$

Place the $v_x$and $v_y$values in the given equation:

$v=\sqrt{v_x^2+v_y^2}$

Therefore,

$v=\sqrt{v_x^2+v_y^2}$

$=\sqrt{(6.0\mathrm{m}/\mathrm{s}{)}^2+(5.33\mathrm{m}/\mathrm{s}{)}^2}$

$=8.03m/s$

The projectile's speed at $t=20s$is $8.02m/s$.

We are given the force in two dimensions in the form

Let's apply the equation of horizontal component to find the time:

$\stackrel{→}{v_x}=\stackrel{→}{v_i}\hat{i}+{∫}_0^t{a}_{x}dt$

${\stackrel{→}{v}}_x=8.0\mathrm{m}/\mathrm{s}\hat{i}+{∫}_0^t(-1.0t\hat{i})dt$

$\stackrel{→}{v_x}=8.0\mathrm{m}/\mathrm{s}\hat{i}+\frac{-1}{2}{t}^2$

$0=8.0\mathrm{m}/\mathrm{s}+\frac{-1}{2}{t}^2$

$8.0\mathrm{m}/\mathrm{s}=\frac{1}{2}{t}^2$

$t=\sqrt{16}$

Simplify,

The projectile moving parallel to the$y-$axis at $4s$.