Question

HELP ~ 50 Points!
A set of keys slides along the floor, coming to rest 4.3 meters from where they started. If the initial speed of the keys was 5.6 m/s then ....

the acceleration of the keys is __________.
The time it takes the keys to stop is __________.

Answer 1

• Distance=s=4.3m
• Initial velocity=u=5.6m/s
• Final velocity=v=0
• Acceleration=a
• time=t

$\\ \sf\longmapsto a=\dfrac{v^2-u^2}{2s}$

$\\ \sf\longmapsto a=\dfrac{-(5.6)^2}{2(4.3)}$

$\\ \sf\longmapsto a=\dfrac{-31.36}{8.6}$

$\\ \sf\longmapsto a=-3.6m/s^2$

And

$\\ \sf\longmapsto v=u+at$

$\\ \sf\longmapsto t=\dfrac{v-u}{a}$

$\\ \sf\longmapsto t=\dfrac{-5.6}{-3.6}$

$\\ \sf\longmapsto t=1.5s$