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.


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.
