Solving Kinematic Movement Equation: Bike Acceleration

[FelixLudi]

Hi,
It's my first post here on the forum and I'm just looking for an answer to a basic kinematic movement equation.

This is the text of a problem:
"A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike."

Now, in the solution, they used v²=v₀²+2*a*s.
(s=d, v=v_f, v₀=v_i)
To derive from that:
v²=2*a*s
v²/2*s=a

(7,1m/s)²/2*35,4m=a
50,41/70,8=a
a=0,712m/s²

Now my question is how I got a different answer by using a different equation, i.e.
a=v/t
a=v/(s/v)
a=7,1/(35.4/7.1)
a=7.1/4,985
a=1.424m/s²

What is the difference/mistake here that I/they did to get a different result?
Thank you in advance.

 

[PeroK]

Now my question is how I got a different answer by using a different equation, i.e.
a=v/t
a=v/(s/v)

In these equations you are using $v$ as final speed $a=v/t$ and as average speed $t=s/v$.

 

[FelixLudi]

But if I use a=vt then I will get m/s * s = m so vt should equal to s instead of a.
By that I mean shouldn't vt be equal to s and not a.

I don't understand what did I do wrong.

Edit: If I use a=vt then a=v*(s/v) which is basically a=s.

 

[PeroK]

But if I use a=vt then I will get m/s * s = m so vt should equal to s instead of a.
By that I mean shouldn't vt be equal to s and not a.

I don't understand what did I do wrong.

Sorry, I had a typo. I meant $a = v/t$, of course.

$vt \ne s$ if $v$ is final speed.

For example. You go out in your car and crawl through traffic for nearly an hour. Finally, you get onto a clear road and accelerate to $100km/h$. So, you say:

My speed is $v = 100km/h$

I've been driving for $t = 1$ hour.

Therefore, the distance I've driven is $s = vt = 100km$

But, you might have only driven a few kms. In any case, you can't use your final speed to calculate how far you have driven.

 

[CWatters]

Now my question is how I got a different answer by using a different equation, i.e.
a=v/t
a=v/(s/v)

You have calculated t from s/v but that only works if v is the average velocity. In this problem v is the final velocity not the average.

In this problem the velocity increases from 0 to v in a straight line (constant acceleration) so the average velocity is v/2 and your equation becomes

a = v/(s/v/2)
= v²/2s

which is the same equation as method 1).

 

[FelixLudi]

Oh, so it has to be average speed instead of final speed?
Didn't even notice that factor.

Thank you very much.