Question

a skater is moving at a speed of 13 m/s when she starts to brake. she comes to a stop after traveling a distance of 8 m. what was her acceleration?

Answer 1

Its a kinematic problem, use the formulas:
\[X _{(t)}=-\frac{ 1 }{ 2 }at ^{2}+v _{0}t+x _{0}\], and \[x_{0}=0m\] then use: \[v=at+ v_{0} \rightarrow t=\frac{ v-v_{0} }{ a}\]
we replace t in the first formula:\[X _{(t)}=-\frac{ 1 }{ 2 }a (\frac{ v-v_{0} }{ a })^{2}+v_{0}.(\frac{ v-v_{0} }{ a }) =-\frac{ 1 }{ 2 }\frac{ (v-v_{0})^{2} }{ a }+v_{0}.\frac{ (v-v_{0}) }{ a}\]now, we take common factor (1/a):
\[X _{(t)}=\frac{ 1 }{ a }[v _{0}(v-v _{0})-\frac{ 1 }{ 2 }.(v-v_{0})^{2}]\rightarrow a=\frac{ [v_{0}.(v-v_{0})-\frac{ 1 }{ 2 }.(v-v _{0})^{2}] }{ X _{(t)} }\]
Finally you just have to replace the values, knowing that:
\[v_{0}=13\frac{ m }{ s }\]\[X_{(t)}=13m\]

Answer 2

but now im checking again the problem i think you need another fact like velocity.

Answer 3

so her acceleration would be 5m/s? I think i did that wrong..

Answer 4

velocity or aceleration is 5m/s?

Answer 5

\[v _{f}^{2}=v _{i}^{2}+2ad\]

Answer 6

Hi welcome to Openstudy!

Since the skater went to stop the final velocity should be 0m/s
use kinematics equation : \(\sf \Large V_f^2=V_i^2+2ad\)

evaluate the equation to solve for acceleration:
\(\sf \Large a=\frac{V_f^2-v_i^2}{2d}\)
plug in your given values and you're done.