to settle an argument regarding a cars acceleration.

calltronics

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A car at rest at Point A accelerates to a Point B, and then it applies the brakes to return to rest at Point C.
Given that: -
· The distance between Point A and Point C is fixed
· The time taken to travel between Point A and Point C is fixed

Solve for constant acceleration and deceleration over the same distance and time: -
1. Min acceleration when the acceleration and deceleration (brake) are equal.
2. Formula for deceleration (brake) for any given acceleration.

Cheers for looking
Steve
 
A car at rest at Point A accelerates to a Point B, and then it applies the brakes to return to rest at Point C.
Given that: -
· The distance between Point A and Point C is fixed
· The time taken to travel between Point A and Point C is fixed

Solve for constant acceleration and deceleration over the same distance and time: -
1. Min acceleration when the acceleration and deceleration (brake) are equal.
2. Formula for deceleration (brake) for any given acceleration.

Cheers for looking
Steve

Start with naming unknowns:

For 1

Let the acceleration = a and deceleration = -a (acceleration and deceleration cannot be equal - their magnitude can be equal).

now continue....

Please share your work with us .

If you are stuck at the beginning tell us and we'll start with the definitions.

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http://www.freemathhelp.com/forum/th...217#post322217
 
Hi Subhotosh Khan

I am really stuck and have gone around in circles I did start with: -

r = Accel Period
f = Brake Period
t = total
S = distance
U = initial velocity
V = final velocity
T = time
a = acceleration

Obviously: -
The known total distance St = Accel distance Sr+ Brake Distance Sf
The known total time Tt = Accel Time Tr + Brake Time Tf
The starting velocity Ur of acceleration phase and the final velocity Vf of the braking phase must be 0
The final accel period velocity Vr must equal the start brake period velocity Uf

Using s = ut + 1/2(at2)

the the total distance = UfTf +1/2 -aTf
2 + UrTr +1/2 aTr2
As Uf and Vr are O (they are at rest)

Total Distance =
1/2 -a(Tt​ -Tr)
2 +UrTr +1/2 aTr2

Then I get lost give me a clue to the next step.
I think I have to replace U in terms of t and a but I am not sure

 
Last edited:
Hi Subhotosh Khan

I am really stuck and have gone around in circles I did start with: -

r = Accel Period
f = Brake Period
t = total
S = distance
U = initial velocity = 0
V = final velocity = 0
T = time....but then you used lower-case t for the time variable
a = acceleration
d = deceleration (not necessarily equal to a)

Obviously: -
The known total distance St = Accel distance Sr+ Brake Distance Sf
The known total time Tt = Accel Time Tr + Brake Time Tf.
The starting velocity Ur of acceleration phase and the final velocity Vf of the braking phase must be 0
The final accel period velocity Vr must equal the start brake period velocity Uf
....Let the Maximum Velocity Vm = Vr = Uf (you only need one variable name)

Using s = ut + 1/2(at2)

the the total distance = UfTf +1/2 -aTf
2 + UrTr +1/2 aTr2
As Uf and Vr are O (they are at rest)

Total Distance =
1/2 -a(Tt​ -Tr)
2 +UrTr +1/2 aTr2

Then I get lost give me a clue to the next step.
I think I have to replace U in terms of t and a but I am not sure

Instead of using distance as a function of t^2, use the equation that change of velocity = acceleration times time:
......Vm = a Tr = - d Tf

Since velocity is a linear function of time, the average during acceleration is Vm/2, and the average during deceleration is also Vm/2. Then given the total time Tt and the total distance St, the average velocity is
......Vm/2 = St / Tt

For case 1, the times Tr = Tf = Tt/2, and a = -d

For case 2, a will be given and you have to solve for d.
 
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