Question

A particle initially located at the origin has an acceleration of a 2.00j m/s2 and an initial velocity of v-6.00i m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). ti+ t2 j) m (b) Find the velocity of the particle at any time t. I + tj) m/s (c) Find the coordinates of the particle at t 5.00 s. (d) Find the speed of the particle at t 5.00 s. m/s

Answer 1

Answer:

• $\vec{r}(t) = (-6.00 \frac{m}{s} \ t , \frac{1}{2} \ 2.00 \ \frac{m}{s^2} \ t^2 )$
• $\vec{v}(t) = (-6.00 \frac{m}{s}, 2.00 \ \frac{m}{s^2} \ t )$
• $\vec{r}( 5.00 \ s) = (-30.00 \ m , 25.00 \ m )$
• $| \vec{v} (5.00 \ s) | =11.66 \frac{m}{s}$

Explanation:

The initial position of the particle, $\vec{r}_0$, is:

$\vec{r}_0 = (0,0)$

the initial velocity is:

$\vec{v}_0 = - 6.00 \ \frac{m}{s} \hat{i} = (-6.00 \frac{m}{s},0)$

and the initial acceleration:

$\vec{a} = 2.00 \ \frac{m}{s^2} \ \hat{j} = ( 0, 2.00 \ \frac{m}{s^2})$

a

The position $\vec{r}$ at time t is

$\vec{r}(t) = \vec{r}_0 + \vec{v}_0 \ t + \frac{1}{2} \ \vec{a} \ t^2$

So, for our problem is:

$\vec{r}(t) = (0,0) + (-6.00 \frac{m}{s},0) \ t + \frac{1}{2} \ ( 0, 2.00 \ \frac{m}{s^2}) \ t^2$

$\vec{r}(t) = (0 -6.00 \frac{m}{s} \ t , 0 + \frac{1}{2} \ 2.00 \ \frac{m}{s^2} \ t^2 )$

$\vec{r}(t) = (-6.00 \frac{m}{s} \ t , \frac{1}{2} \ 2.00 \ \frac{m}{s^2} \ t^2 )$

b

The velocity $\vec{v}$ at time t is

$\vec{v}(t) = \vec{v}_0 + \vec{a} \ t$

So, for our problem is:

$\vec{v}(t) = (-6.00 \frac{m}{s},0) + ( 0, 2.00 \ \frac{m}{s^2}) \ t$

$\vec{v}(t) = (-6.00 \frac{m}{s}, 2.00 \ \frac{m}{s^2} \ t )$

c

At time 5.00 seconds the position will be:

$\vec{r}( 5.00 \ s) = (-6.00 \frac{m}{s} \ 5.00 \ s , \frac{1}{2} \ 2.00 \ \frac{m}{s^2} \ (5.00 \ s ) ^2 )$

$\vec{r}( 5.00 \ s) = (-30.00 \ m , 25.00 \ m )$

d

and the speed will be :

$| \vec{v} (5.00 \ s) | = |(-6.00 \frac{m}{s}, 2.00 \ \frac{m}{s^2} \ 5.00 \ s) |$

$| \vec{v} (5.00 \ s) | = |(-6.00 \frac{m}{s}, 10.00 \ \frac{m}{s}) |$

$| \vec{v} (5.00 \ s) | = \sqrt{ (-6.00 \frac{m}{s})^2 + (10.00 \ \frac{m}{s}))^2 }$

$| \vec{v} (5.00 \ s) | = \sqrt{ 36.00 \frac{m^2}{s^2} + 100.00 \ \frac{m^2}{s^2}}$

$| \vec{v} (5.00 \ s) | =11.66 \frac{m}{s}$