A particle move rectilinearly with a constant acceleration of 1m/s

^{2}.its speed after 10second is 5 m/s .The distance cover by particle in this duration is ? ( intial and final velocities are in opposite direction )-
CREATE FREE ACCOUNT
- Access thousands of free resources from expert tutors
- Comment on posts and interact with the authors
- Ask questions and get free answers from tutors
- View videos, take interactive quizzes, and more!

- Become a Student
- Become a Student
- Sign In

Tutors, please sign in to answer this question.

If it started from 0 and accelerated at 1 m/s/s for 10 s how fast would it be going?

=> 10 m/s

So it must have started from a negative speed of -5 m/s to get to 5 m/s after 10 sec.

Are you sure it is asking for distance and not displacement?

Start at a point, go for 5 sec negative while slowing to 0 m/s,

then 5 sec positive speeding up to 5 m/s

And you get back to where you started => displacement is zero.

The distance traveled is from the formula x_{f} = x_{0} + v_{0}t + 1/2(at^{2})

but you need to do it in 2 parts

SO

x = 1/2(at^{2}) = 1/2(1 m/s/s) (5 s)^{2}

= 12.5 m in negative direction

them another 12.5 meters coming back

= 25 meters distance traveled.

Jordan E.

Professional Cellist, Experienced Cello Instructor

New Brunswick, NJ

4.6
(13 ratings)

Jim D.

Patient and Creative Tutor for English, Writing, and Public Speaking

Woodbridge, NJ

Brian G.

Math tutor for high school and college students

Highland Park, NJ

4.9
(15 ratings)