Finding the max speed of an object between two point

[Mark Hogan]

Hello everybody,

I am trying to calculate the maximum velocity(x) an object reaches between two point. The object accelerates from point A to a certain velocity(x) and then decelerates to point B. The acceleration and deceleration are known but are different. The distance from point A to point B is known. Is there a formula I can use to calculate the maximum velocity between the two points?

I have attempted to solve this by graphing the distance and time of the object and I have attempted rearranging formulas to do this but with no luck.

Your help would be greatly appreciated,
Mark

 

[jbriggs444]

You will need some additional information before being able to solve this. As written, it is possible that the object accelerates from A almost all the way to B and then decelerates for only an instant before zipping past point B. Or that the object starts from A with non-zero velocity, accelerates momentarily and then decelerates all the way to a stop at point B.

Is it also true that the object starts at rest at point A and ends at rest at point B?

 

[CWatters]

Ok so it...

Starts from rest and accelerates at "a" to velocity "v" over distance s₁
Then it decelerates at "d" from "v" to rest over distance s₂

General form of suvat equation...

v² = u² + 2as

where

v is the final velocity
u = initial velocity
a = acceleration or deceleration
s = displacement

In this problem we can write the equation twice, once for the bit while accelerating and one decelerating..

Accelerating..
v² = 2as₁

Decelerating..
0 = v² + 2(-d)s₂

We also know that

s₁ + s₂ = Total distance

I believe these can be solved to find "v".

 

[Mark Hogan]

Ok so it...

Starts from rest and accelerates at "a" to velocity "v" over distance s₁
Then it decelerates at "d" from "v" to rest over distance s₂

General form of suvat equation...

v² = u² + 2as

where

v is the final velocity
u = initial velocity
a = acceleration or deceleration
s = displacement

In this problem we can write the equation twice, once for the bit while accelerating and one decelerating..

Accelerating..
v² = 2as₁

Decelerating..
0 = v² + 2(-d)s₂

We also know that

s₁ + s₂ = Total distance

I believe these can be solved to find "v".

Hello jbriggs444 and CWatters,

Thank you for your help. I may not have explained my problem well.

 

[Mark Hogan]

have

I will explain it using an example:
So an object starts from rest. It accelerates at 2000mm/s2 then decelerates at 3000mm/s2 to rest. The distance from start to finish let's say 200mm. But is it possible to calculate the max velocity reached while traveling between the points?

 

[BruceW]

yes, it is possible to find the max velocity using just this information. But it does take a few steps. I recommend drawing out a graph of space against time, (and possibly the derivates too ). And try to split the problem into two - the acceleration section and the deceleration section. You can use the time of the change from accel. to decel. as a variable to solve for.

 

[CWatters]

I will explain it using an example:
So an object starts from rest. It accelerates at 2000mm/s2 then decelerates at 3000mm/s2 to rest. The distance from start to finish let's say 200mm. But is it possible to calculate the max velocity reached while traveling between the points?

That's exactly the situation I was addressing in post #3 above. You just need to solve the three equations I gave to find "v". Not too hard.

 

[Mark Hogan]

Hello everybody,

Thanks for your help. I found a solution. I used the following formulas -> (time = √2Xdistance / accel) and the formula -> (velocity final = time x accel x velocity initial)

I then graphed the velocity for both acceleration and deceleration vs the distance traveled for a number of points. Where the two curves on the graphs met gave me the
max velocity reached between the two points.

Thanks again for your help,
Mark

 

[Khashishi]

Did you check your result? Your method sounds a little suspicious.

 

[CWatters]

velocity final = time x accel x velocity initial

That's not correct. I think you mean ..

velocity final = (time x accel) + velocity initial

PS The answer I got was about 0.7 m/s

 

[Chestermiller]

Did you check your result? Your method sounds a little suspicious.

I agree. I would like to see them do the problem using t₁ and t₂ for the two times, and then compare the results with the other method. I think the answer will come out differently. It also seems easier to do this problem using the times.

Chet

 

[Mark Hogan]

Hello everybody,

Sorry I meant... velocity final = (time x accel) + velocity initial... (typing error!). CWatters I also came out with 0.7m/s. I graphed the max velocity for both acceleration and deceleration against different distances in increments of 100mm for 1m. I was able to read from this the max velocity reached between the points. I have included a picture of my graph for you to see.

Mark

 

[CWatters]

I did it by solving the three equations I posted earlier. Only takes a few steps. That gives the equation...

V² = 2*D*a*d/(a+d)

where
V = max velocity
D = Distance from A to B (0.2m)
a = acceleration (2m/s/s)
d = deceleration (3m/s/s)

Plug in the numbers and you get 0.7m/s

 

[CWatters]

I agree. I would like to see them do the problem using t1 and t2 for the two times, and then compare the results with the other method. I think the answer will come out differently. It also seems easier to do this problem using the times.

I can't see how to do that as you don't know the total time. I think you would have to eliminate time as one of the steps in solving the equations.

 

[CWatters]

Here is my working...

The three equations I wrote above were:

Accelerating phase..
v² = 2as₁......(1)

Decelerating phase..
0 = v² + 2(-d)s₂.....(2)

and
s₁ + s₂ = D (the total distance)... (3)

Rearrange 1) and 2) to give

s₁ = V²/2a
s₂= V²/2d

Adding together (see eqn 3) gives

V²/2a + V²/2d = D

Rearrange to give

V² = 2*D*a*d/(a+d)

 

[Mark Hogan]

Here is my working...

The three equations I wrote above were:

Accelerating phase..
v² = 2as₁......(1)

Decelerating phase..
0 = v² + 2(-d)s₂.....(2)

and
s₁ + s₂ = D (the total distance)... (3)

Rearrange 1) and 2) to give

s₁ = V²/2a
s₂= V²/2d

Adding together (see eqn 3) gives

V²/2a + V²/2d = D

Rearrange to give

V² = 2*D*a*d/(a+d)

Hi CWatters, Thanks for your help this is a much better way of doing it :-)

 

[BruceW]

yeah. suvat is a lot faster here. I would have solved the equations without suvat, but it just depends which way you prefer to do it. (I have long forgotten suvat, so I prefer not to use it).
p.s. nice graph. graphs are always good :)