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I have a vehicle traveling at initial speed $100\:\mathrm{m/s}$. Expected destination speed is also $100\:\mathrm{m/s}$. Maximum deceleration is $3$. Maximum acceleration is $10$. Distance to the target is $2000\:\mathrm{m}$. The vehicle should reach the target in 40s.

The vehicle should follow a linear path, decelerate($a_1$) to a certain speed($v_1$) and accelerate back($a_2$) to reach the target precisely on $40\:\mathrm{s}$.

Note: $v_1 > 0$, $a_1 < 3$, $a_2 < 10$.

How do we find $a_1$, $a_2$ and $v_1$ ??

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Actually $a_1 \ge -3$ –  Henry Apr 27 '11 at 11:46
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3 Answers 3

The units of a1 and a2 are m/(s^2)? Gravities? I will assume the former.

If you define: t1 = time of acceleration a1 (with 0>a1>-3); t2 = time of acceleration a2 (t2 = 0 at start of acceleration a2); and restrict the solution space to t1+t2=40 (no dead time between a1 and a2). Then you can get a quadratic to describe motion from start to intermediate point, and another quadratic to describe motion from the intermediate point to 2000m. The intermediate point is where you transition from a1 to a2. That is 3 equations and 4 unknowns. When I do the substitution (no guarantee I got it right) I get a quadratic in t1:

a(t1^2) + b(t1) + c = 0; where

a = 1

b = -80

c = (4000+1600(a2))/(a2-a1)

This is a solution space which gets you to 2000m at t1+t2 = 40 seconds. The next thing you need to do is ensure you get to 2000m with a velocity of 100m/s. Since there is not dead time, and since v(t=0s) = v(t=40s) = 100m/s; then (t1)(a1) + (t2)(a2) = 0. This eqn yields a1 = -(a2)(t2)/(t1) = -(a2)(40-t1)/(t1). substitute that back into the quadratic.

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In general, Henry's answer is correct - this task has no solution to satisfy all the conditions.

The normal set of equations would be:

$v_1=100, v_3=100, t=40, s=2000, a1<3, a2<10$

$s_1 = v_1 t_1$ - first idle part at max speed.

$s_2 = v_1 t_2 - a_1 t_2^2/2$ - deceleration part

$s_3 = v_2 t_3 + a_2 t_3^2/2$ - acceleration part

$v_2 = v_1 - a_1*t_2$ - the minimum speed

$v_2 = v_3 - a_2*t_3$ - we must accelerate to end speed.

$s_1 + s_2 + s_3 = s$ - total distance = 2000m

$t_1 + t_2 + t_3 = t$ - total time = 40s

$t_1 >= 0, t_2 >= 0, t_3 >= 0, s_1 >= 0, s_2 >= 0, s_3 >= 0$

$a_1 >= 0, a_2 >= 0, a_1 <= 3, a_2 <= 10, v_2>0$

If you want, try to solve it fully. You'll eventually arrive at a quadric equation with negative delta. If you want a simplified solution, assume $a_1 = 3, a_2 = 10, t_0=0$

The $v_2$ equations give $t_3= 0.3 t_2$; $t_2 + t_3 = 40$. You can calculate the results of this time split yourself.

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This problem looks likely to be badly posed: either there are a large number of solutions or there are none. Only in the most artificial cases will there be a single solution at an extreme: either do not decelerate or accelerate at all so you go as far as possible in the time, or decelerate and then accelerate as hard as possible all the time to go as short a distance as possible in the time.

Let's take the hard deceleration and acceleration possibility. Since the final speed is the same as the initial speed, the changeover time must be at about time 30.769 seconds. This means $v_1$ is about 7.692 m/s and the distance travelled to the changeover point is about 1656.805 metres; so at 40 seconds the speed is back to 100 m/s and the further distance travelled is about 497.041 metres, making a total of well over 2000 metres.

Too far.

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