Get the minimum taken to move from A to B given a cap on Velocity

[thejackal]

Homework Statement

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

Homework 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

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

 

[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.

 

[thejackal]

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.

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.

 

[haruspex]

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.

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

 

[HalfLostAndSearching]

To find the minimum time taken to cover the distance x, we can use the equation x = V0t + 0.5at^2, where V0 is the initial velocity and a is the acceleration. However, we must also take into account the maximum permissible velocity, Vmax. This means that the object cannot exceed Vmax at any point during its motion.

To account for this, we can use the equation V = V0 + at, where V is the velocity at time t. We can set this equal to Vmax and solve for t, giving us the time at which the object reaches its maximum velocity. We can then plug this value of t into the equation x = V0t + 0.5at^2 to find the distance traveled at this time.

If this distance is less than x, then the object has not yet reached its destination and we need to continue the calculation. If the distance is equal to x, then this is the minimum time taken to cover the distance.

If the distance is greater than x, then we need to adjust the value of Vmax and recalculate until we find the correct value. This is because the object cannot exceed Vmax, so we need to find the maximum velocity that will still allow the object to reach the distance x.

Overall, to find the minimum time taken to cover the distance x, we need to iteratively adjust the value of Vmax until the distance traveled at the time when V = Vmax is equal to x. This will give us the minimum time taken to cover the distance x while still adhering to the maximum velocity constraint.