Car braking problem: finding speed after going 1/3 of stopping distance

[jhson114]

A car traveling at a speed of vo = 59 m/s stops smoothly (that is, its deceleration is constant) over a distance of d = 166 m.

There are 3 parts to this quetions.
1. what is its acceleration during the time it is stopping
2. how long does it take for the car to stop
3. after the car has gone 1/3 of the stopping distance what is its speed

i have answered the first two questions, but i can't answer the last question.
i got -10.5 m/s^2 and 5.6 seconds for first two answer. to get third answer i simply did Vf = Vi + A*deltaT which came out Vf = 59-10.5(5.6/3) = 39.4
first two answers are correct but the third one is wrong. can someone please tell me what I'm doing wrong. thank you

 

[Leong]

You cannot divide 5.6/3 just because the distance is one-third of d because we have an acceleration in this case. instead, use $$v^2=u^2+2as$$.

 

[jhson114]

but since the acceleration is constant, i thought i could do it that way.

 

[jhson114]

what is "u" and "s"? v is velocity, a is acceleration

 

[arildno]

Ok, you are assuming that the time taken to travel a third of the distance equals a third of the time taken to travel the whole distance.
This is incorrect; that assumes a linear relationship between time passed and distance travelled.
What formulas do you think might help you?

 

[jhson114]

the v2=u2+2as probably helps, but i don't know what u and s stands for.

 

[jhson114]

okay. i figured it out. thanks guys :)