PROBLEM: A train accelerates from rest at one train station with $a= 0.72t$ $m/s^2$, until it reaches its maximum speed of $18 m/s$. Its deceleration heading into the next train station is $3.0 m/s^2$. If the journey between the two stations takes 6 minutes and 20 seconds, what is the distance between the stations?
WORKED ANSWER: I initially drew a $v-t$ graph, with maximum velocity of $18 m/s$ and maximum time of $380$ seconds. So the graph looks like a triangle with a peak of $18$. One edge of the triangle has a slope of $0.72t$ and the other edge has a slope of $-3$. So, the triangle is halved at time $t_1$ which I presumed is unknown.
PERSONAL QUESTION 1/2: According to the problem given, did I draw the graph correctly in order for me to answer the question?
WORKED ANSWER continued:
For $t \le t_1$ $v=0.36t^2$:
$18 = 0.36t_1^2$
$t_1=5\sqrt2$
$\therefore$ $\Delta x=A_1+A_2$
$=(1/2*5\sqrt2*18)+(1/2*18*(380-5\sqrt2))$
$=3420m$
PERSONAL QUESTION 2/2: If the graph I have drawn is correct, is my procedure correct and hence is my answer correct?
Best Answer
Your graph for the first part isn’t correct. The integral of a linear function gives a parabola.
Just like how uniformly increasing velocity (constant acceleration) produces a parabolic $x-t$ (position-time) graph. For $a = 0.72t$, $j = 0.72$. $j$ represents “jerk” (rate of change of acceleration).
Hence, $$\Delta x = \frac{1}{6}jt^3+\frac{1}{2}a_0t^2+v_0t = \frac{1}{6}(0.72)\left(5\sqrt{2}\right)^3 = 30\sqrt{2}$$
(Since the object starts from rest, the terms involving $a_0$ and $v_0$ become $0$.) The formula above can be derived via repeated integration.
You can also use integration, which is also viable:
$$v(t) = \int a(t)dt = \int 0.72t = \frac{1}{2}0.72t^2 = 0.36t^2$$
You have $t_0 = 0$ and $t_1 = 5\sqrt{2}$, so
$$\Delta x = \int_{0}^{5\sqrt{2}}0.36t^2 = \frac{1}{3}(0.36)\left(5\sqrt{2}\right)^3 = 30\sqrt{2}$$
The rest of your answer is correct because there is constant acceleration, producing a triangular $v-t$ graph.