# Homework Help: Problems with acceleration results established from equations of linear motion

1. Jul 5, 2012

### Agrech

I have 3 separate cases of an object decelerating at different rates. The aim is to find the acceleration for each case.
Case 1
Initial velocity: 1.663m/s
Displacement: 0.12m
Time: 0.139s

Case 2
Initial velocity: 1.663m/s
Displacement: 0.12m
Time: 0.155s

Case 3
Initial Velocity: 1.663m/s
Displacement: 0.12m
Time: 0.24s

The attempt at a solution
Simply using the equations of linear motion (s=ut+1/2at^2) I got the following results:
Case 1: -11.5063m/s^2
Case 2: -11.4685m/s^2
Case 3: -9.69167m/s^2

Why? I expected the magnitude of deceleration to increase from case 1 to case 3
Also deceleration of 11m/s^2 seems excessively high. I need to understand this for an assignment due in a few days.
I've done this many times now and with different methods such as finding the final velocity first and then finding the acceleration but every method gives me the same results or at least the same pattern of the deceleration decreasing.

Last edited: Jul 5, 2012
2. Jul 5, 2012

### janny

Hi,
imagine yourself sitting in a car moving with the given initial velocity and then applying the brakes. You do that three times. First time you need 0.139s, second time 0.155s and third time 0.24s. At which of the cases have you applied the brakes hard, harder, the hardest?
About feeling the 11m/s^2 excessively high: Compare it with g, the acc. due to gravity.

3. Jul 5, 2012

### azizlwl

Simply using the equations of linear motion (s=ut+1/2at^2) I got the following results
-----------------------------------------

Since the displacement and initial velocity are constant, increasing value of t should be followed with decreasing value of a.

4. Jul 5, 2012

### lewando

In janny's scenario, the moving car will travel 0.12m in the shortest time if no brakes are applied. The longest time will result with the largest deceleration.

5. Jul 5, 2012

### Agrech

@janny, Thankyou for your help. So far this is exactly my thoughts, in case 1 the brakes are applied hard, case 2 harder and case 3 the hardest. This is why I expected that case 3 would the highest magnitude of deceleration, however I consistently get case 3 being the lowest.
Regarding your comment about comparing the acceleration with gravity, that is exactly why I feel that is too high.

6. Jul 5, 2012

### Agrech

I agree with this but my results aren't showing that.

7. Jul 5, 2012

### Agrech

@lewando yes my scenario is very similar.
I arrived at my results by using s=ut+1/2at^2
Rearranged to a=2(s-ut)/t^2
And the results are in my original post

8. Jul 5, 2012

### janny

Agrech,

You have a sign failure with your calculations.
Its rather s=ut-1/2 at2 than s=ut+1/2 at2.

Last edited: Jul 5, 2012
9. Jul 5, 2012

### janny

PS: Or as you suggested you use s=ut+1/2 at2 but bearing in mind that this would give -a.

Last edited: Jul 5, 2012
10. Jul 5, 2012

### lewando

Ponder this for a while: http://fooplot.com/plot/h238j2y69v

For this family of equations, no amount of deceleration is going to get to the 0.12m mark in 0.24s on the first encounter.

Maybe you can describe in more detail your experiment.

11. Jul 5, 2012

### PhanthomJay

As one can see, the use and interpretation of plus and minus signs in Intro Physics are cause for much confusion. There is nothing wrong with the answers listed in the original post. The magnitude of the deceleration is the greatest when the time is the shortest, that is, when the brakes are applied the hardest. The 1 g deceleration is not too bad as long as you are buckled up for safety.

12. Jul 5, 2012

### lewando

PhantomJay, in all three cases, the initial velocities are the same and the distance traversed is the same. I would re-assert that the shortest time occurs when there is zero deceleration.

As you decelerate while traversing the distance, you slow down and the time should increase accordingly.

At some large enough deceleration, you will begin to move backwards and not even make the 0.12m stretch.

I believe these statements to be correct, but I have been wrong before.

13. Jul 6, 2012

### janny

lewando,

As you decelerate while traversing the distance, you slow down and the time should increase accordingly.

Though thats exactly what I have thought so far, I must contradict, PhantomJay is right, I give it a trial prooving that mathematically:

the traveled distance in each cases is constant, i.e

ut1 +1/2 a1t12 = ut2 +1/2 a2t22
t1 +1/2 a1t12/u = t2 +1/2 a2t22/u
Now, let t2>t1 i.e there is a z from ℝ such that t2=zt1, hence

t1 +1/2 a1t12/u = zt1 +1/2 a2(zt1)2/u ⇔
zt1-t1 = 1/2 a1t12/u - 1/2 a2(zt1)2/u ⇔
zt1-t1 = a1t12-a2z2t12/2u ⇔
2u(zt1-t1 ) = t12(a1-a2z2) ⇔
2u(zt1-t1)/t12 = a1-a2z2.
Now, since 2u(zt1-t1)/t12 > 0 it follows that a1-a2z2 also must be > 0. But that in return implies a1>a2z2 and so a1>a2. Alltogether we see: t1<t2 => a1>a2. That is, the shorter the time the greater the magnitude of acceleration, as PhanthomJay suggested.

PS: a should be read as magnitude of a, for I couldnt find the magnitude dashes I have omited them.

14. Jul 6, 2012

### lewando

When using the kinematic equation s = ut + 1/2at2, to eliminate confusion, realize this:

"a" is not a vector.
"a" is not the magnitude of a vector.
"a" is a scalar.

The value of the scalar "a" is determined by the x or y axis acceleration vector that the equation is describing.

An acceleration of 10m/s2 in the positive x-axis direction would result in the scalar value of 10m/s2.
An acceleration of 10m/s2 in the negative x-axis direction would result in the scalar value of -10m/s2.

The terms acceleration and deceleration can also be a source of confusion. Recommend avoiding the term deceleration for now.

That being said, @janny, your analysis is correct. The interpretation of your result needs some thought.

For objects traveling in the +x direction with identical initial velocities as they enter a fixed distance interval, what you said is true:
Using actual scalar values:
a1 = 10m/s2 => t1 (shorter than t2, because of larger final velocity)
a2 = 5m/s2 => t2

In this case t1<t2 and a1>a2, in agreement with your analysis.

Using another set of scalar values:
a3 = -5m/s2 => t3
a4 = -10m/s2 => t4 (longer than t3, because of smaller final velocity)

In this case t3<t4 and a3>a4, also in agreement with your analysis.

15. Jul 6, 2012

### lewando

@Agrech--I hope you are still following. How certain are you of you initial velocity? 1.663m/s looks very precise and consistent. Was this measured or calculated? Reason I ask--if you recompute your accelerations based on an initial velocity of 1m/s, the results make more sense.

16. Jul 6, 2012

### PhanthomJay

The SUVAT equations tell no lies. Barring a math error, the original answers are correct. The negative value of the acceleration implies that if the object is initially moving to the right, the acceleration is to the left. The paradox of why the shorter times yield higher values of the leftward accelerations is solved by realizing that there are two roots for the time in the quadratic equation; when both roots are positive, the given times at a positive displacement of 0.12 m may very well be occurring when the object is moving backwards to the left, having earlier already passed that point when moving right.

Last edited: Jul 6, 2012