Calculate speed and acceleration

[lamefeed]

A person runs 100 m in 11.8 s. She had constant acceleration the first 3.0 s and after that she had constant speed(velocity if you will)

s=vt
v = v0 + at
s= v0t + 1/2at^2
2as = v^2 – Vo^2

I have tried a bunch of times but I seem to get it wrong every single time. This has given me a small headache the last couple of hours.

$s(100)=\frac{1}{2}*a*3^2*t$
$a=\frac{100}{9/2*11.8}$

Hope anyone can make sense out of this and give me a explanation on how to get the correct answers.

The correct answers is in the spoiler

Spoiler

a = 3.24m/s^2
v=9.71m/s

 

[Chestermiller]

If a was her acceleration during the first 3 sec, what was her speed and distance after those three seconds (in terms of a)? If her speed remained constant during the next 8.8 sec, what was her additional distance during those 8.8 sec(in terms of a)? In terms of a, what was her total distance?

 

[lamefeed]

If a was her acceleration during the first 3 sec, what was her speed and distance after those three seconds (in terms of a)? If her speed remained constant during the next 8.8 sec, what was her additional distance during those 8.8 sec(in terms of a)? In terms of a, what was her total distance?

Total distance is 100 m, she ran 100 m in 11.8 s, she had constant acceleration during the first 3 s after that(the remaining 8.8 s) she had constant v.

$s = vt$
$s = \frac{1}{2}at^2$
A combination of the two is what I suspect to be the correct formula

something along the lines of this:

$s = \frac{1}{2}at_1^2 + vt_2$

Where t(1) = 3 s, and t(2) = 8.8 s

 

[Chestermiller]

Good. Now all you need to do is substitute the relationship for v as a function of a and t₁. (v doesn't change after the initial 3 seconds). This will give you what you need to solve for a.

Chet