Question

In a particular Cartesian coordinate system, the y and z-components of the acceleration are zero and the x-component varies as given by the following function: ax(t) = 9t - 6t2 + 25e-t/F, where t is in seconds, ax is in meters per square second, and the constant F is in seconds. At t = 0, the particle was at position x = 7 m with a velocity pointing towards the positive x-axis and having magnitude 15 m/s. In the following problem you can take the constant F = 1.0 s.

Answer 1

a) The velocities of the particle are $v_{x}(1) \approx 33.303\,\frac{m}{s}$, $v_{x}(2) \approx 38.617\,\frac{m}{s}$ and $v_{x}(3) \approx 25.255\,\frac{m}{s}$, respectively.

b) The accelerations of the particle are $x(1) \approx 32.197\,m$, $x(2) \approx 69.383\,m$ and $x(3)\approx 103.245\,m$, respectively.

Note - Statement is incomplete, complete description is presented below:

In a particular Cartesian coordinate system, the y and z-components of the acceleration are zero and the x-component varies as given by the following function: $a_{x}(t) = 9\cdot t -6\cdot t^{2} + 25\cdot e^{-\frac{t}{F} }$, where $t$ is in seconds, $a_{x}$ is in meters per square second, and the constant $F$ is in seconds. At $t = 0$, the particle was at position $x = 7\,m$ with a velocity pointing towards the positive x-axis and having magnitude $15\,\frac{m}{s}$. In the following problem you can take the constant $F = 1.0\,s$.

(a) Find the instantaneous velocity, in meters per second, at $t = 1\,s$, and $t = 2\,s$, and $t = 3\,s$.

(b) Find the position, in meters, of the particle at $t = 1\,s$, $t = 2\,s$ and $t = 3\,s$.

The functions velocity and position of the particle are found by integration of the function acceleration in time. That is to say:

$v_{x}(t) = \int {a_{x}(t)} \, dt$

$v_{x}(t) = \int {(9\cdot t -6\cdot t^{2}+25\cdot e^{-\frac{t}{F} })} \, dt$

$v_{x}(t) = 9\int {t} \, dt - 6\int {t^{2}} \, dt + 25\int {e^{-\frac{t}{F} }} \, dt$

$v_{x}(t) = \frac{9}{2}\cdot t^{2}-2\cdot t^{3}-25\cdot F\cdot e^{-\frac{t}{F} } + C_{1}$ (1)

$x(t) = \int {v_{x}(t)} \, dt$

$x(t) = \int {\left(\frac{9}{2}\cdot t^{2}-2\cdot t^{3} - 25\cdot F\cdot e^{-\frac{t}{F} } + C_{1} \right)} \, dt$

$x(t) = \frac{9}{2}\int {t^{2}} \, dt - 2\int {t^{3}} \, dt -25\cdot F \int {e^{-\frac{t}{F} }} \, dt + C_{1} \int \, dt$

$x(t) = \frac{3}{2}\cdot t^{3}-\frac{1}{2}\cdot t^{4}+25\cdot F^{2}\cdot e^{-\frac{t}{F}} + C_{1}\cdot t + C_{2}$ (2)

Where $C_{1}$ and $C_{2}$ are integration constants.

If we know that $t = 0\,s$, $F = 1\,s$, $v_{x} = 15\,\frac{m}{s}$ and $x = 7\,m$, then the integration constants are, respectively:

$C_{1} = 40$, $C_{2} = -18$

Now we evaluate each function at given instants:

a) Velocities:

(t = 1 s)

$v_{x}(1) = \frac{9}{2}\cdot 1^{2}-2\cdot 1^{3}-25\cdot e^{-1} + 40$

$v_{x}(1) \approx 33.303\,\frac{m}{s}$

(t = 2 s)

$v_{x}(2) = \frac{9}{2}\cdot 2^{2}-2\cdot 2^{3}-25\cdot e^{-2} + 40$

$v_{x}(2) \approx 38.617\,\frac{m}{s}$

(t = 3 s)

$v_{x}(3) = \frac{9}{2}\cdot 3^{2}-2\cdot 3^{3}-25\cdot e^{-3} + 40$

$v_{x}(3) \approx 25.255\,\frac{m}{s}$

b) Acceleration:

(t = 1 s)

$x(1) = \frac{3}{2}\cdot 1^{3}-\frac{1}{2}\cdot 1^{4}+25\cdot e^{-1} + 40\cdot 1 - 18$

$x(1) \approx 32.197\,m$

(t = 2 s)

$x(2) = \frac{3}{2}\cdot 2^{3}-\frac{1}{2}\cdot 2^{4}+25\cdot e^{-2} + 40\cdot 2 - 18$

$x(2) \approx 69.383\,m$

(t = 3 s)

$x(3) = \frac{3}{2}\cdot 3^{3}-\frac{1}{2}\cdot 3^{4}+25\cdot e^{-3} + 40\cdot 3 - 18$

$x(3)\approx 103.245\,m$

The velocities of the particle are $v_{x}(1) \approx 33.303\,\frac{m}{s}$, $v_{x}(2) \approx 38.617\,\frac{m}{s}$ and $v_{x}(3) \approx 25.255\,\frac{m}{s}$, respectively.

The accelerations of the particle are $x(1) \approx 32.197\,m$, $x(2) \approx 69.383\,m$ and $x(3)\approx 103.245\,m$, respectively.

We kindly invite to see this question on integrals: brainly.com/question/23567529