Respuesta :
Answer:
- [tex]\vec{r}(t) = (-6.00 \frac{m}{s} \ t , \frac{1}{2} \ 2.00 \ \frac{m}{s^2} \ t^2 )[/tex]
- [tex]\vec{v}(t) = (-6.00 \frac{m}{s}, 2.00 \ \frac{m}{s^2} \ t )[/tex]
- [tex]\vec{r}( 5.00 \ s) = (-30.00 \ m , 25.00 \ m )[/tex]
- [tex]| \vec{v} (5.00 \ s) | =11.66 \frac{m}{s}[/tex]
Explanation:
The initial position of the particle, [tex]\vec{r}_0[/tex], is:
[tex]\vec{r}_0 = (0,0)[/tex]
the initial velocity is:
[tex]\vec{v}_0 = - 6.00 \ \frac{m}{s} \hat{i} = (-6.00 \frac{m}{s},0)[/tex]
and the initial acceleration:
[tex]\vec{a} = 2.00 \ \frac{m}{s^2} \ \hat{j} = ( 0, 2.00 \ \frac{m}{s^2})[/tex]
a
The position [tex]\vec{r}[/tex] at time t is
[tex]\vec{r}(t) = \vec{r}_0 + \vec{v}_0 \ t + \frac{1}{2} \ \vec{a} \ t^2[/tex]
So, for our problem is:
[tex]\vec{r}(t) = (0,0) + (-6.00 \frac{m}{s},0) \ t + \frac{1}{2} \ ( 0, 2.00 \ \frac{m}{s^2}) \ t^2[/tex]
[tex]\vec{r}(t) = (0 -6.00 \frac{m}{s} \ t , 0 + \frac{1}{2} \ 2.00 \ \frac{m}{s^2} \ t^2 )[/tex]
[tex]\vec{r}(t) = (-6.00 \frac{m}{s} \ t , \frac{1}{2} \ 2.00 \ \frac{m}{s^2} \ t^2 )[/tex]
b
The velocity [tex]\vec{v}[/tex] at time t is
[tex]\vec{v}(t) = \vec{v}_0 + \vec{a} \ t[/tex]
So, for our problem is:
[tex]\vec{v}(t) = (-6.00 \frac{m}{s},0) + ( 0, 2.00 \ \frac{m}{s^2}) \ t[/tex]
[tex]\vec{v}(t) = (-6.00 \frac{m}{s}, 2.00 \ \frac{m}{s^2} \ t )[/tex]
c
At time 5.00 seconds the position will be:
[tex]\vec{r}( 5.00 \ s) = (-6.00 \frac{m}{s} \ 5.00 \ s , \frac{1}{2} \ 2.00 \ \frac{m}{s^2} \ (5.00 \ s ) ^2 )[/tex]
[tex]\vec{r}( 5.00 \ s) = (-30.00 \ m , 25.00 \ m )[/tex]
d
and the speed will be :
[tex]| \vec{v} (5.00 \ s) | = |(-6.00 \frac{m}{s}, 2.00 \ \frac{m}{s^2} \ 5.00 \ s) |[/tex]
[tex]| \vec{v} (5.00 \ s) | = |(-6.00 \frac{m}{s}, 10.00 \ \frac{m}{s}) |[/tex]
[tex]| \vec{v} (5.00 \ s) | = \sqrt{ (-6.00 \frac{m}{s})^2 + (10.00 \ \frac{m}{s}))^2 }[/tex]
[tex]| \vec{v} (5.00 \ s) | = \sqrt{ 36.00 \frac{m^2}{s^2} + 100.00 \ \frac{m^2}{s^2}}[/tex]
[tex]| \vec{v} (5.00 \ s) | =11.66 \frac{m}{s}[/tex]