First, let's assume the object starts with a velocity vi=0, and furthermore for the first part let's also assume a uniform acceleration. I'm also assuming you are asking this for a non-calculus based course.
Part 1:
Over 10 seconds, the velocity increases from vi=0 to vf=10 m/s, or in other words, each second the velocity increases by 1 m/s. This is an acceleration of a=1 m/s/s. Using the kinetics equation:
vf=vi+at
10=0+a(10)
a=1
Part 2:
The distance traveled can be found
using:
d=vi+vf2t
d=102(1)=50
Or, alternatively:
d=vit+12at2
d=0(1)+12(1)(102)
d=50
Explanation:
sana naka help, pag nasagutan mo na pala sorry po, pa brainliest po;(
Answers & Comments
Answer:
First, let's assume the object starts with a velocity vi=0, and furthermore for the first part let's also assume a uniform acceleration. I'm also assuming you are asking this for a non-calculus based course.
Part 1:
Over 10 seconds, the velocity increases from vi=0 to vf=10 m/s, or in other words, each second the velocity increases by 1 m/s. This is an acceleration of a=1 m/s/s. Using the kinetics equation:
vf=vi+at
10=0+a(10)
a=1
Part 2:
The distance traveled can be found
using:
d=vi+vf2t
d=102(1)=50
Or, alternatively:
d=vit+12at2
d=0(1)+12(1)(102)
d=50
Explanation:
sana naka help, pag nasagutan mo na pala sorry po, pa brainliest po;(