How to calculate traveled distance with non constant acceleration in time?

Question

I know this formula D = vt + 1/2at^2* for calculating the distance given initial velocity, time and acceleration. But what if my acceleration is not static, but increasing exponentially defined by f(n) = a^n? If the acceleration a starts increasing at distance x and time tx (reference point a) and I do a measurement after n seconds (reference point b) with the formula above, the results would be incorrect, because the acceleration initially was smaller.

Example: at reference point a object travels at velocity 10 m/s and acceleration is 2 m/s^2 and time is 40 seconds (tx). After this point acceleration starts to increase by a(n) = a^n where n is every second after tx. So at n=2 (total 42 seconds) seconds acceleration would be 4 m/s^2, n=3 it would be 8 m/s^2 and so on. I want to measure the distance from a to b, where b is after n=10 (total 50 seconds).

Example2: at reference point a object travels at velocity 10 m/s and acceleration is 2 m/s^2 and time is 40 seconds (tx). After this point acceleration starts to increase by a(n) = a^dn where dn is every meter after a. So at dn=2 (total a+2 meters) seconds acceleration would be 4 m/s^2, dn=3 it would be 8 m/s^2 and so on. I want to measure the distance from a to b, where b is after after 10 seconds (total 50 seconds).

Answer 1

Use calculus: $$v(t)\equiv\int a(t)\ dt$$ and $$x(t)\equiv\int v(t)\ dt.$$

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update:

First try some special cases:

• $a(t)=(0{\rm\ m/s^2})$
• $a(t)=a_0$ (where $a_0$ is a constant)
• $a(t)=j_0 t$
• $a(t)=A\sin(\omega t)$

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(By the way, "acceleration is not static, but increasing exponentially" as "$f(n) = a^n$" doesn't make sense to me.)

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example: With $a(t)=j_0 t$,
then $v(t)=\frac{1}{2}j_0 t^2 + v_0$
and $x(t)=\frac{1}{6}j_0 t^3 +v_0 t+x_0$.

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Update2

You need to write down an acceleration function as a possibly piecewise function of t. ( Draw a graph of $a(t)$. )

Once you have that function, use the formula to get the velocity and then the next formula to get the position.