How do I determine the time it takes to accelerate if I only know the distance traveled and what the amount of acceleration is? I am currently losing as I don't really have any clue on how to solve this calc problem as there is nothing we covered in class like this:
Determine how many seconds it would take for a car to accelerate uniformly from 0 to 60 miles per
hour using $\frac{1}{20}$th of a mile long track. Give your answer in seconds.
What I've determined so far is that I somehow need to find the rate of acceleration, and then use integration to find the number of seconds (this could be wrong, I honestly don't know). What I'm thinking this problem looks like is something like this:
$$\int_0^{60}a(t)dt = \frac{1}{20}$$
To my understanding what I wrote here says that from 0 to 60 miles per hour, $\frac{1}{20}$th of a mile has been traveled. If I can somehow determine $a(t)$ then I can somehow find $t$. Sorry if this isn't making a lot of sense, I'm trying to best to show what my thought process is on this problem, but I really don't know how to solve something like this, it feels like I'm missing a lot of information.
In case someone was curious I am in Calculus I in my first year of college.
Also, just to be clear, I'm not really asking for an answer to the question, maybe just some insight on how I can approach it as I'm not sure my thinking is correct.
 A: You're on the right track (no pun intended)! The integral you wrote is with respect to $dt$, so the bounds are time values. You know two pieces of information, the final speed (60 mph) and the final distance (0.05 miles). The integral of acceleration with respect to time is a velocity, and the integral of velocity with respect to time is a distance. You don't know the time, so let's call it $T$. Because you start from rest, you know that:
$$\int_0^Ta(t)dt=v(T)-v(0)=60\ \mathrm{mph}-0\ \mathrm{mph}=60\ \mathrm{mph}$$
You also know that your initial position is 0, so:
$$\int_0^T v(t)dt=x(T)-x(0)=0.05\ \mathrm{miles}-0\ \mathrm{miles}=0.05\ \mathrm{miles}$$
Finally, you know your acceleration is uniform, so $a(t)=a$, for some constant $a$. Therefore, you have two unknowns $a$ and $T$, and two equations. This should let you solve for everything.
A: Unfortunately, your integral of
$$\int_0^{60}a(t)dt = \frac{1}{20} \tag{1}\label{eq1A}$$
is not quite correct, although you have the right general idea. Note the period of time of acceleration is not $60$, as this is the final speed instead. Also, the integral of acceleration is the velocity, not the distance, i.e., $\frac{1}{20}$.
Instead, note the phrase "accelerate uniformly" means the acceleration is a constant, call it $k$. This means
$$a(t) = k \tag{2}\label{eq2A}$$
Using that $v(0) = 0$, then integrating \eqref{eq2A} to get the velocity $v(t)$ gives
$$v(t) = \int_{0}^{t}a(x)dx = kt \tag{3}\label{eq3A}$$
Also, since we've been given the distance travelled, assign the position $s(0) = 0$ so that $s(t)$ is the distance travelled. Integrating \eqref{eq3A} then gives
$$s(t) = \int_{0}^{t}v(x)dx = \frac{kt^2}{2} \tag{4}\label{eq4A}$$
Let $t_f$ be the final time, in hours, when the car has reached $60$ mph. Using this in \eqref{eq3A} gives
$$v(t_f) = kt_f = 60 \tag{5}\label{eq5A}$$
At that time, the distance travelled becomes $\frac{1}{20}$ miles. Using this in \eqref{eq4A} gives
$$s(t_f) = \frac{kt_f^2}{2} = \frac{1}{20} \tag{6}\label{eq6A}$$
Next, \eqref{eq6A} divided by \eqref{eq5A} gives $t_f$ in hours. This then needs to be converted to seconds. I'm leaving it for you to do these last couple of steps.
A: Alt. hint (no calculus required): constant acceleration means the speed varies linearly, which in turn means the distance covered over a period of time is the average speed multiplied by the time.
In this case, the distance is $\dfrac{1}{20} \,\text{mile}$, the average speed is $\dfrac{0 + 60}{2} = 30 \,\text{mph}$, and dividing the two gets the elapsed time.
A: With uniform acceleration you don't need calculus, beyond that used to integrate constant acceleration $a$, giving:
$$v = v_0 + at$$ and $$x = x_0 + v_0t+\frac{1}{2}at^2$$ which combine to $$v^2 = v_0^2+2a(x-x_0)$$
$x_0$ is starting position, which you can set to $0$, and $v_0$ is the starting velocity, which you know is $0$.
A: 
maybe just some insight on how I can approach it as I'm not sure my thinking is correct.

A general hint for all problems: always include the units in your calculations.
E.g. the given integration can be seen as blatantly wrong even if you use only the units without any values:

*

*Left Side = [acceleration × time] = [m/h/s × s] = [m/h], which is velocity.

*Right Side = [m], which is distance.

Regardless of the actual numbers and calculations, velocity can't be equal to distance, so something isn't right.
Similarly the integral has:

*

*Upper value unit is [m/h], which is velocity.

*The "dt" has [s] as the unit, which is time.

Velocity can't be equal to distance, so again something is obviously wrong.
A: 
Here is a more Physics based approach.
In problems like these, I often like to think in terms of graphs.
The image is of a graph of Velocity versus time , where Velocity is in mph and time in Hours.
Let's say it took the car $t$ hours to accelerate. As we know acceleration is constant, graph of Velocity versus time will be a straight line (why?). Also, we know at $t$ = 0, $v$ = 0 ($v$ represents velocity). After $t$ hours, $v = 60$.
Now, what is the total displacement? We know it is the area under the curve!
So, $\frac{1}{20}$ miles = $\frac{1}{2} \cdot t$ (hours) $\cdot 60$ $\frac{miles}{hour}$
So, we get $t = \frac{1}{600}$ hours.
I will let you convert that to seconds.
