Distance travelled during acceleration

[stupidlawyer]

Homework Statement

i am a stupid criminal lawyer, i have a driving case where a client drove 80 metres from a stopped position. if client accelerates at rate of 0-100km/h in 7 seconds how fast will he be traveling when he gets to the 80th metre?

Homework Equations

i know nothing

The Attempt at a Solution

please help, my client is innocent.

 

[Qube]

1) List your knowns - displacement and initial velocity - and calculate acceleration

2) Find relevant equation that has the variables V(final), d, and a.

3) Plug in and solve

Here's my work:

[PLAIN]http://i.min.us/icDaju.jpg

 

[SteamKing]

The units of vf^2 should be (m/s)^2.

 

[Rayquesto]

You want to know the average acceleration first right? so, then your simple answer is 100/7 meters/second^2=about 14.29 meters/second^2. The client's initial velocity is 0 meters per second. so, v= 0 plus the (time it take to travel 80 meters times 100/7 m/s^2). to calculate time, you got to say (80meters=1/2at^2). Therefor, it took about 3.35 seconds to reach 80 meters. Ultimiteley then if you plug in 3.35seconds times 14.29= velocity after 80 meters. So, your answer is 48meters/second, unless the acceleration was inconsistent. that should be it though. correct me if I am wrong, because I'm just a student. remember 48 meters/second after 80 meters if the acceleration was constant, which I think is.

 

[Rayquesto]

oh oopse sorry! i forgot to convert hours to seconds! my bad!

 

[Rayquesto]

in that case, the acceleration is 3.96m/s^2 and so it would be 80=(1/2)(3.96m/s^2)(t)^2

time=6.356seconds
vf=(6.356s)(3.96m/s^2)=25.17meters/second
the guy who first replied was completely right then thanks man!

 

[Qube]

Uh ... no - that's not how it works. Simply because vf is squared doesn't mean to arbitrarily square part of the other side.

The units of vf^2 should be (m/s)^2.

 

[willem2]

Uh ... no - that's not how it works. Simply because vf is squared doesn't mean to arbitrarily square part of the other side.

The units must match. If they don't it means you have forgotten to non-arbitrarily square the other side.

$v_i^2$ and 2ad both have units of $m^2s^{-2}$

 

[willem2]

in that case, the acceleration is 3.96m/s^2 and so it would be 80=(1/2)(3.96m/s^2)(t)^2

time=6.356seconds
vf=(6.356s)(3.96m/s^2)=25.17meters/second
the guy who first replied was completely right then thanks man!

This is correct, but assumes that there was constant acceleration. If the car accelerates faster at the start and slower later on (wich is likely, because it must change to a higher gear at some point), it will reach the 80m point earlier and have less speed.

 

[Rayquesto]

So, are you suppposed to use the vf^2=vi^2 plus 2acceleration times distance equation if you're not sure if acceleration is constant?

 

[ideasrule]

So, are you suppposed to use the vf^2=vi^2 plus 2acceleration times distance equation if you're not sure if acceleration is constant?

That equation also assumes acceleration is constant. If it's not, the problem is unsolvable unless you know precisely how acceleration changes with time.