1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Motion Problem .

  1. Jul 12, 2015 #1
    1. The problem statement, all variables and given/known data

    A car travelling with constant acceleration increases its speed from 10 m/s to 50 m/s over a distance of 60 m . how long does this take?

    a. 2 seconds b. 4 seconds .

    3. The attempt at a solution
    Data :
    Avg Velocity = 10 + 50 / 2 = 30m/s
    Distance = 60 meter
    time = ?

    Solution :

    S= V x T

    making "T" subject . Then

    T= S/V
    T= 60/30

    T = 2 seconds . This Answer is correct , but please tell me that the way I have solved this is correct or not? Thanks
     
  2. jcsd
  3. Jul 12, 2015 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your solution is fine.
     
  4. Jul 12, 2015 #3

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I agree with Doc Al. Your method is correct.

    But I'm in a nit picky mood this morning. Use parenthesis in the expression for calculating the average velocity in order to show that the 10 and 50 are added together first before dividing by 2:

    ( 10 + 50 ) / 2.

    Anyway, good work!
     
  5. Jul 12, 2015 #4
    Thanks , But I am not satisfied with my solution, u know why sir? because S = vt is possible when initial velocity is 0 means herz equation s = vit + 1/2 a t^2 , but this eq can only be reduced to s =vt when acceleration is unity and initial velocity is 0 , in my case it isnt ? please make me clear or I am going fine?
     
  6. Jul 12, 2015 #5
    Check this:
    We have three equations of motion for constant acceleration along 1D,
    1. v=u+at,
    2. v*v - u*u=2as, and
    3. S=ut + 1/2at*t

    okay

    Here u(initial velocity)=10m/s, v(final velocity)=50m/s, a=constant (so apply equations as it is a 1D problem), s=60, t=?
    Using eq. 2 we get
    2500-100=2*a*60
    a=20m/s*s

    Now using this value of 'a' in eq 1 we get;

    50=10+20*t
    t=2s

    Thats it fella...
     
  7. Jul 12, 2015 #6

    Yes Thanks , this was really a perfect solution , I was wondering how s = vt gonna solve this , Thanks alot
     
  8. Jul 12, 2015 #7

    Doc Al

    User Avatar

    Staff: Mentor

    What you're using, whether you realize it or not, is ##\Delta S = v_{ave}t##. Which is fine here.
     
  9. Jul 12, 2015 #8
    Thanks , but the sir who replied above gave this solution.

    Here u(initial velocity)=10m/s, v(final velocity)=50m/s, a=constant (so apply equations as it is a 1D problem), s=60, t=?
    Using eq. 2 we get
    2500-100=2*a*60
    a=20m/s*s

    Now using this value of 'a' in eq 1 we get;

    50=10+20*t
    t=2s


    I think it was simple nasic problem, I followed question wrong i think.
     

    Attached Files:

  10. Jul 12, 2015 #9
    Top tip:Whenever there is constant acc in 1D go for the equations...
     
  11. Jul 12, 2015 #10
    Its very kind of you , Thanks for sharing knowledge , actually I am going to give medical college admission test this year , it has 3 subjects on which test will be based , 1 BIology . 2 Physics 3 Chemistry . so these little concepts will help me alot , Thanks
     
  12. Jul 12, 2015 #11

    Doc Al

    User Avatar

    Staff: Mentor

    There's nothing wrong with this solution, of course. But note the extra steps involved.

    Your original approach using average velocity was just fine.
     
  13. Jul 12, 2015 #12
    Yes I understood Sir , Thanks , I will keep all these things in my knowledge . Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Motion Problem .
  1. Motion problem (Replies: 1)

  2. Problems on motion (Replies: 1)

  3. Motion Problem (Replies: 3)

  4. Problem on motions (Replies: 1)

  5. Motion problem (Replies: 5)

Loading...