# Find max velocity given initial velocity, final velocity, acceleration and time

I am trying to find the maximum velocity in a trapezoidal-like motion profile given the initial velocity, the final velocity, the acceleration, the total time and distance traveled.

If the initial and final velocities were zero, I would simply use this formula:

$$v=-\frac{a\left(-t\pm\sqrt{\frac{at^2-4d}{a}}\right)}{2}$$

where $$a$$ = acceleration, $$t$$ = time and $$d$$ = distance

I have tried to draw the trapezoidal-like shape and add all the different parts of the geometric figure like this:

$$d = d_{a} + d_{vmax} + d_{d}$$

Distance traveled during acceleration: $$d_{a} = v_0t_a + \frac{1}{2}at_a^2$$

Distance traveled during deceleration: $$d_{d} = v_1t_d + \frac{1}{2}at_d^2$$

Distance traveled with maximum velocity: $$d_{vmax} = v(t - t_a - t_d)$$

Then I substitute $$t_a$$ and $$t_d$$ using $$t\:=\:\frac{v-v_0}{a}$$, resulting in the following equation:

$$d\:=\:v\left(\frac{x-v}{a}\right)+\frac{1}{2}a\left(\frac{x-v}{a}\right)^2\:+x\left(t-\left(\frac{x-v}{a}\right)-\left(\frac{x-w}{a}\right)\right)\:+\:w\left(\frac{x-w}{a}\right)+\frac{1}{2}a\left(\frac{x-w}{a}\right)^2$$

where $$x$$ = the unknown maximum velocity, $$v$$ = initial velocity, $$w$$ = final velocity.

Obviously this does not work as intended. The results I get don't make much sense.

I would really appreciate some help on this matter.

• See this question and answer. Conservation of energy is your friend. – amd Oct 3 '18 at 0:58
• Is the deceleration equal in magnitude to the acceleration? Why does it have a positive sign? – Andrei Oct 4 '18 at 14:12