# Homework Help: Get the minimum taken to move from A to B given a cap on Velocity

1. Apr 23, 2014

### thejackal

1. The problem statement, all variables and given/known data
Find the minimum time taken for the object to cover the distance x

I'm given the following:
x = the distance i need to cover
V0 = the initial velocity which is zero
a = the maximum permissible acceleration
Vmax = the maximum permissible velocity

2. Relevant equations

the speed of the object is (v0 + t * a) at time t
the object will be (v0 * t + 0.5 * a * t^2) away

3. The attempt at a solution

I used the kinematic equation x/2 = V0 + 0.5at^2 to find t. In some case this worked for example using the test case x = 1, a = 2, Vmax = 10 then t = 1.414...; x = 1, a = 1, Vmax = 1 then t = 2s
but for the case x = 10, a=1, Vmax = 1 I failed to get the correct answer (11 s). Would anyone here please help me understand how to account for Vmax and a. I'm not sure how the equations given in the problem statement aid in getting the answer. thanks

2. Apr 23, 2014

### haruspex

I don't understand the /2 in "x/2" in your equation. Neither do I see how you got t=sqrt(2) for x =1, a=2, with or without the /2. Did you mean x=1, a=1?
For this sort of problem, the answer is not going to pop out directly from the equations. You have to break it into cases. First, suppose Vmax is not reached, and solve with SUVAT as you have. Then check whether the final velocity exceeds Vmax. If so, start again, finding the time taken to reach Vmax. See how you go from there.

3. Apr 23, 2014

### thejackal

I used x/2 because the object after reaching halfway would have to stop accelerating in order to reach the final destination (exactly 1m) with a velocity of 0 m/s. so if x = 1 then using half of x i can find the time it took to reach halfway then multiply by 2 to get the total time for the entire distance.

4. Apr 23, 2014

### haruspex

You didn't mention that requirement originally. Is it a requirement? Anyway, what I suggested for taking Vmax into account should work either way.