Finding the max speed of an object between two point

  • #1

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
jbriggs444
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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?
 
  • #3
CWatters
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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".
 
  • #4
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.
 
  • #5
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 lets say 200mm. But is it possible to calculate the max velocity reached while traveling between the points?
 
  • #6
BruceW
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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.
 
  • #7
CWatters
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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 lets 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.
 
  • #8
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 travelled 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
 
  • #9
Khashishi
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Did you check your result? Your method sounds a little suspicious.
 
  • #10
CWatters
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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
 
  • #11
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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
 
  • #12
Capture.PNG
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
 
  • #13
CWatters
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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
 
  • #14
CWatters
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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.
 
  • #15
CWatters
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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)
 
  • #16
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 :-)
 
  • #17
BruceW
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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 :)
 

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