Need help for a mechanic problem

[mad]

Hi guys.. this is my first post. Sorry if I use terms that you don't use since I translate it from french..

I need help doing this problem. I tried it a few times but I can't find the answer. I tried to do the V(x) (speed or velocity I think) and find it from there but still can't manage to have it.. so if someone could help me.. this is the problem:

A car at rest accelerates uniformly for 200m. She then continues at constant speed for 160m and decelerates 50meters before stopping. The entire movement lasts 33seconds.

Question: how much time was she at constant speed ?

Sorry for my english..
thanks for the help

 

[The Bob]

I believe you need another variable. You have the distance, you want to find the time but what is the speed?

Acceleration(m/s²) = Change in Velocity(m/s) / Time Taken for Change (s)

or a = (v-u)/t

This is one equation you are likely to need.

The other is:

Speed(m/s) = Distance Travelled(m)/Time Taken(s)

or s = d/t

There we go for equations.

The Bob (2004 ©)

 

[mad]

I believe you need another variable. You have the distance, you want to find the time but what is the speed?

Acceleration(m/s²) = Change in Velocity(m/s) / Time Taken for Change (s)

or a = (v-u)/t

This is one equation you are likely to need.

The other is:

Speed(m/s) = Distance Travelled(m)/Time Taken(s)

or s = d/t

There we go for equations.

The Bob (2004 ©)

Thanks, but without the time taken, I can't use those 2 formulas.. ill try and post later.
thanks

 

[arildno]

This can be dealt with in the following manner:
1. Let the speed at the constant level be V
2. In the period of uniform accelerations, the average velocity (distance traveled/time spent) is V/2
I'll prove that if you like..
3.Now let t1 be the time spent in acceleration, t2 the time spent at constant speed, and t3 the time spent in deceleration.
Hence:
V/2*t1=200
V*t2=160
V/2*t3=50
Or, by division:
t1=2*(200/160)t2=5/2t2
t3=2*(50/160)t2=5/8t2

Or, since t1+t2+t3=33, we get:
33/8*t2=33, that is t2=8

 

[mad]

This can be dealt with in the following manner:
1. Let the speed at the constant level be V
2. In the period of uniform accelerations, the average velocity (distance traveled/time spent) is V/2
I'll prove that if you like..
3.Now let t1 be the time spent in acceleration, t2 the time spent at constant speed, and t3 the time spent in deceleration.
Hence:
V/2*t1=200
V*t2=160
V/2*t3=50
Or, by division:
t1=2*(200/160)t2=5/2t2
t3=2*(50/160)t2=5/8t2

Or, since t1+t2+t3=33, we get:
33/8*t2=33, that is t2=8

Wow, thanks a lot! How are you able to do that so easily? Do you have a trick or something ;)

I just have a question: why is the average velocity v/2 ? (.2)
Velocity = distance/time.
You know it traveled 200m but don't know the time it spent.. could you explain me how you found v/2 ?

thank you

 

[arildno]

Wow, thanks a lot! How are you able to do that so easily? Do you have a trick or something ;)

I'm old, but not too senile yet

I'll argue for the average value as follows, when we start initially at rest:
1. Having some constant acceleration "a", we know that the velocity v(t) at time "t" can be written as:
v=at
2. We also know that the distance traveled r(t) is: r= at^{2}/2

3.Rewriting, we have:
r=(at/2)t=(v/2)t
4. Hence, the average velocity r/t equals v/2; that is, half the final velocity.

 

[The Bob]

I'll argue..................half the final velocity.

This is what I was trying to get at but I couldn't. I feel quite useless now. Oh well... that is life. The young give way to the older way of thinking.

The Bob (2004 ©)

 

[arildno]

This is what I was trying to get at but I couldn't. I feel quite useless now. Oh well... that is life. The young give way to the older way of thinking.

The Bob (2004 ©)

But only for a short time..very soon I'll descend past you into senility and decrepitude.

 

[The Bob]

But only for a short time..very soon I'll descend past you into senility and decrepitude.

Not ncessarily. Just live well.

The Bob (2004 ©)

 

[arildno]

Thx!
You seem to be a promising young man with possibly a bright future in front of you (aargh, that one should be in "Lame Jokes")

 

[The Bob]

Thx!
You seem to be a promising young man with possibly a bright future in front of you (aargh, that one should be in "Lame Jokes")

The Bob (2004 ©)