# A particle is moving along a straight line and the acceleration is given by a(t)=3t+5, (units m/s2). a)Find v(t) , the velocity at time t if the initial velocity is v(0) = -4 ft/s? b)Find the total distance traveled during the time interval [0, 2]?

Jan 28, 2018

Part a): $v \left(t\right) = \frac{3}{2} {t}^{2} + 5 t + 1.219$ ($\text{m/s}$)
Part b): $\Delta x = 16.438$ $\text{m}$

#### Explanation:

Note: I apologize in advance for the long solution; I am not sure how much one who reads this understands and have attempted to give some explanation for each step taken. Hope this helps!

Let's begin by writing out what we are given:
$a \left(t\right) = 3 t + 5$ (${\text{m/s}}^{2}$)
$v \left(0\right) = - 4$ ($\text{ft/s}$)

Before we begin to solve the problem, we need to convert acceleration and velocity into some common unit. In this solution, we will be using SI ($\text{m}$, $\text{m/s}$, ${\text{m/s}}^{2}$)

Since $1$ $\text{m}$ is equal to $3.281$ $\text{ft}$, we can use some simple dimensional analysis to convert $v \left(0\right)$ from $\text{ft/s}$ into $\text{m/s}$:
v(0)=(-4" ft")/"s"*(1 " m")/(3.281" ft")=1.219 " m/s"

Remember that $a = \frac{\mathrm{dv}}{\mathrm{dt}}$, so:
$\mathrm{dv} = a \mathrm{dt}$

Integrating both sides, we have:
$\int \mathrm{dv} = \int a \mathrm{dt}$

So,
$v = \int a \mathrm{dt}$

Applying the above formula, we can find $v \left(t\right)$ in units of $\text{m/s}$:
$v \left(t\right) = \int \left(3 t + 5\right) \mathrm{dt}$

Here, we must remember how to integrate a polynomial:
$\int \left(k {x}^{n}\right) \mathrm{dx} = k \frac{{x}^{n + 1}}{n + 1} + C$

Since $a \left(t\right) = 3 t + 5$ is a polynomial, we can apply this integration method, so:
$v \left(t\right) = \int \left(3 t + 5\right) \mathrm{dt} = \frac{3}{2} {t}^{2} + 5 t + C$

Since we are given that $v \left(0\right) = 1.219$ $\text{m/s}$, we can solve for $C$:
$v \left(0\right) = \frac{3}{2} {\left(0\right)}^{2} + 5 \left(0\right) + C = 1.219$

Therefore,
$C = 1.219$ and the answer to part a) is $v \left(t\right) = \frac{3}{2} {t}^{2} + 5 t + 1.219$ $\text{m/s}$

To answer part b), we use the $v \left(t\right)$ equation found in part a) to find the point(s) at which the particle stops and turns around by setting $v \left(t\right) = 0$ and solving for the roots of the quadratic

However, we can take a shortcut and notice that in the quadratic, both the $t$ coefficient and constant are positive, which would thus result in only negative roots. Therefore, we do not need to worry about solving for $t$'s at which $v \left(t\right) = 0$ since we are only being asked to find the total distance traveled from $t \in \left[0 , 2\right]$

Here, we need to remember that $v = \frac{\mathrm{dx}}{\mathrm{dt}}$, solving for $x$ by a similar process as we did for $v$ earlier:
$v = \frac{\mathrm{dx}}{\mathrm{dt}}$
$\mathrm{dx} = v \mathrm{dt}$
$\int \mathrm{dx} = \int v \mathrm{dt}$
$x = \int v \mathrm{dt}$

Since we have that $v \left(t\right) = \frac{3}{2} {t}^{2} + 5 t + 1.219$, we an substitute and solve the definite integral:
$\Delta x = {\int}_{0}^{2} \left(\frac{3}{2} {t}^{2} + 5 t + 1.219\right) \mathrm{dt}$
Deltax=(3/6t^3+5/2t^2+1.219t)]_0^2
Deltax=(1/2t^3+5/2t^2+1.219t)]_0^2

Using the Fundamental Theorem of Calculus, we have that:
$\Delta x = \left(\frac{1}{2} {\left(2\right)}^{3} + \frac{5}{2} {\left(2\right)}^{2} + 1.219 \left(2\right)\right) - \left(\frac{1}{2} {\left(0\right)}^{3} + \frac{5}{2} {\left(0\right)}^{2} + 1.219 \left(0\right)\right)$

Plugging this into a calculator we have:
$\Delta x = 16.438$ $\text{m}$