Question

A $2.0kg$projectile with initial velocity $\overrightarrow{v}=8.0\hat{ı}\mathrm{m}/\mathrm{s}$experiences the variable force $\overrightarrow{F}=-2.0t\hat{l}+4.0t^2\hat{ȷ}\mathrm{N}$, where $t$ is in $s$.

a. What is the projectile’s speed at $t=2.0$s?

b. At what instant of time is the projectile moving parallel to the$y-$axis?

Short answer

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

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

Step-by-step solution

Given Information (Part a)

We are given the force in two dimensions in the form

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

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

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

Explanation (Part a)

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,

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

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

Calculate the final velocity (Part a)

Let's compute the final velocity as follow:

$\overrightarrow{v_f}={\overrightarrow{v}}_i+{\int }_0^{t}adt$

For the horizontal component:

$\overrightarrow{v_x}={\overrightarrow{v}}_i\widehat{i}+{\int }_0^t adt$

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

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

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

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

Calculate for the vertical component (Part a)

For the vertical component:

${\overrightarrow{v}}_y={\overrightarrow{v}}_i+{\int }_0^t{a}_{y}dt$

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

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

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

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

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

Therefore,

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

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

$=8.03m/s$

Final Answer (Part a)

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

Given Information (Part b)

We are given the force in two dimensions in the form

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

Explanation (Part b)

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

$\overrightarrow{v_x}=\overrightarrow{v_i}\hat{i}+{\int }_0^t{a}_{x}dt$

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

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

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

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

$t=\sqrt{16}$

Simplify,

$t=4s$

Final Answer (Part b)

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