I faced a kinematics problem in certain physics book today but I thought of asking my doubt here as it is more of a mathematical problem.
"A fun drive in an amusement park runs between two spots that are $2 km$ apart.For safety reasons the acceleration of drive is limited to $+4m/s^2$ or $-4m/s^2$.The maximum permissible magnitude of jerk can be $+1m/s^3$ or $-1m/s^3$.The drive has a maximum speed of $144 km/h$.Find the shortest time taken by the drive to travel between the spots.
I tried to assume the function of displacement as a cubic function of the form $x=t^3+at^2+bt+c $ since jerk (rate change of acceleration ) is possible.But I haven't been able to reach the result still.Can someone guide me please.THANKS.
It seems that you need to compose a piecewise continuous function. If we integrate for constant jerk, $$s=s_0+v_0t+\frac12a_0t^2+\frac16jt^3$$ Where $s_0$, $v_0$, and $a_0$ are the initial displacement, velocity, and acceleration, and $j$ is the constant jerk. So you want to get going with maximum jerk until you reach maximum acceleration, continue with this acceleration for a little while, then with maximum negative jerk decreases your acceleration until it reached $0$ just when you reach maximum velocity. Then cruise along until you get near the end when you have to reverse the process. I am pretty sure that the time to the halfway point should be half the total time.
For the first piece, $j=1$, $a=a_0+jt=t=a_{\max}=4$ when $t=4$ seconds. Then $v=v_0+a_0t+\frac12jt^2=0+0+\frac12\cdot4^2=8$ meters per second, and $s=\frac{32}3$ meters.
For the third piece, recall that maximum velocity is $40$ meters per second $v=40=v_0+a_0t+\frac12jt^2=v_0+4\cdot4-\frac12\cdot4^2=v_0+8$ because it takes $4$ seconds to slow down from maximum acceleration to zero at maximum jerk. Thus we have for the second piece $v=32=v_0+a_0t+\frac12jt^2=8+4t+0$ so the second piece lasts for $6$ seconds. The second piece advances $s$ by $s-s_0=v_0t+\frac12a_0t^2+\frac16jt^3=8\cdot6+\frac12\cdot4\cdot6^2+0=120$ meters, and the third piece advances $s$ by $s-s_0=v_0t+\frac12a_0t^2+\frac16jt^3=32\cdot4+\frac12\cdot4\cdot4^2-\frac16\cdot4^3=\frac{448}3$ meters, so at $t=4+6+4=14$ seconds, we are at $s=\frac{32}3+120+\frac{448}3=280$ meters and cruising at maximum velocity of $40$ meters per second.
It takes another $\frac{1000-280}{40}=18$ seconds to reach the halfway point, so we should complete our journey in $2(14+18)=64$ seconds total.
EDIT: Here is a table of displacement, velocity, and acceleration at the time points when the jerk is changed and the jerk before and after the change. $$\begin{array}{cccccc} t&s&v&a&j_{before}&j_{after}\\ 0&0&0&0&0&1\\ 4&\frac{32}3&8&4&1&0\\ 10&\frac{392}3&32&4&0&-1\\ 14&280&40&0&-1&0\\ 50&1720&40&0&0&-1\\ 54&\frac{5608}3&32&-4&-1&0\\ 60&\frac{5968}3&8&-4&0&1\\ 64&2000&0&0&1&0 \end{array}$$