Acceleration and distance problem

[LinkinPark98]

Homework Statement

A car is driving at a speed of 72km/h. It starts to slow down and it stops after 10s. How long is the trail of braking ?

Homework Equations

t - time
a - acceleration
v0- starting speed
v - final speed

The Attempt at a Solution

So in my book it says that if a object is slowing down the acceleration is negative and the formula for acceleration is a=v-v0\t, assuming that v0=72km/h=20m/s and v=0 , the acceleration is..:
a = -2m/s^2

So if the acceleration is negative then the distance s= v0t - at^2/2..
After some calculating I get that s=300m... But in my book there are solutions and it says that a = 2m/s^2 and that s = 100m... I don´t get it. Please help I need it today.. Thanks :D
Btw. sorry for my bad English :P

 

[Simon Bridge]

When doing kinematics, it is often best to just draw the velocity-time graph.

In your case it starts out at u=72kmph = 20m/s and drops to v=0 in time T=10s.

The area under the graph is the displacement - the graph is a triangle with a base of 10s and a height of 20m/s so the displacement is (20x10)/2=100m.

By the same method: the acceleration is the slope of the graph - which, reading it off the graph, is a=(0-20)/10 = -2m/s/s

If you always draw the v-t graph, you don't need to remember all those equations.

 

[LawrenceC]

" s= v0t - at^2/2.."

s = v0*t + at^2/2

where a = -2.0 m/s

 

[LinkinPark98]

When doing kinematics, it is often best to just draw the velocity-time graph.

In your case it starts out at u=72kmph = 20m/s and drops to v=0 in time T=10s.

The area under the graph is the displacement - the graph is a triangle with a base of 10s and a height of 20m/s so the displacement is (20x10)/2=100m.

By the same method: the acceleration is the slope of the graph - which, reading it off the graph, is a=(0-20)/10 = -2m/s/s

If you always draw the v-t graph, you don't need to remember all those equations.

Hm.. I didn´t (still) learn that triangle graph. I think I understood you but I need a... mathematic way to do this.

" s= v0t - at^2/2.."

s = v0*t + at^2/2

where a = -2.0 m/s

If you mean that instead of "-" I use "+" then why in my book says that if a<0 I use the "-" formula...

 

[Nytik]

You're counting the minus sign twice, since you changed the sign in the formula but then still took 'a' as -2. You only do one or the other.

 

[LinkinPark98]

You're counting the minus sign twice, since you changed the sign in the formula but then still took 'a' as -2. You only do one or the other.

Oh lol. Now I understand... Thank you very much :D

 

[Simon Bridge]

LawrenceC is using the wrong equation (sorry Lawrence) though only because it's the one you were trying to use. He's right about where you went wrong.
You start out knowing initial and final velocities and the time period. You want to know the distance. The kinematic equation for this is:

s=(u+v)T/2

u=20m/s
v=0m/s
T=10s

you don't need the acceleration.

the formula you were trying to use was:

s=uT+aTT/2 for acceleration and

s=uT-aTT/2 for deceleration - which is a no no because it is easy to get confused.

Think of what the equation means - if there was no deceleration, the distance covered would have been just uT, but it has to be less than that because the car was slowing down.

this is why you draw the graph - you don't get confused because you make the equations from the physics instead of getting your physics from the equations.

 

[LinkinPark98]

LawrenceC is using the wrong equation (sorry Lawrence) though only because it's the one you were trying to use. He's right about where you went wrong.
You start out knowing initial and final velocities and the time period. You want to know the distance. The kinematic equation for this is:

s=(u+v)T/2

u=20m/s
v=0m/s
T=10s

you don't need the acceleration.

the formula you were trying to use was:

s=uT+aTT/2 for acceleration and

s=uT-aTT/2 for deceleration - which is a no no because it is easy to get confused.

Think of what the equation means - if there was no deceleration, the distance covered would have been just uT, but it has to be less than that because the car was slowing down.

this is why you draw the graph - you don't get confused because you make the equations from the physics instead of getting your physics from the equations.

OK.. thank you. Guess I need to go the safe way. Once again thanks to everybody :D

 

[LawrenceC]

"LawrenceC is using the wrong equation..."

The above statement is an incorrect statement. Please do not mislead students who are trying to learn. Students have a choice of what applicable equation to use.

s = V0*t + 0.5*a*t^2 = (V0 + Vfinal)*t/2

where V0 is intitial velocity

 

[LinkinPark98]

@LawrenceC- guess you are right because sometimes to hardest formula for other classmates is the easiest for me so I just use the equation I like :D

 

[Simon Bridge]

Lawrence - OK, fair enough, I should rephrase:

The standard approach taught to students for solving problems in kinematics is to select which of the five kinematic equations they have committed to memory is the closest fit to the problem to hand. The equation the student had chosen was not the closest because it requires further algebra.

So I should have said that the choice of equation was "less than ideal".

In my defense I'd point out that the student in question did not become confused. The comment was carefully calculated to cut through the confusion already present and it put the last nail in - though Nytik actually beat me to the job. Sometimes we have to take these small hits to our pride.

I'm sure, however, that LinkinPark98 would appreciate the simple algebra to go from the formula you presented to the one he needed:

s=ut+att/2 : but a=(v-u)/t so
s=ut + ((v-u)/t)tt/2 = ut + (v-u)t/2 : expand the brackets
s=ut + vt/2 - ut/2 =ut/2 + vt/2 : to give
s=(u+v)t/2
I'm sure we agree this is neither difficult nor confusing.
He was still better off with the geometry. vis:

since v=0
s=ut/2

... which is also the equation for the area of a triangle: fancy that! Could have done it in one step.