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Total time travelled

  1. Apr 14, 2009 #1
    1. The problem statement, all variables and given/known data
    a bus starts from rest and accelerating at 1.5m per second squared until it reaches a velocity of 9m/s.The bus continues at this velocity and then decelerates at -2m per second squared until it comes to a stop. 400m from its starting point. How much time did the bus take to cover the 400 m?

    2. Relevant equations

    s =vt, v=u+at, s= ut+ 1/2 * at^2, v^2= u^2 + 2as

    3. The attempt at a solution

    intially found out the time travelled until it reaches 9m/s. so starting from rest, u=0, substituting in 2nd eq. I got t= 9sec. After this I am totally confused. I found out the initial distance travelled, and final diatance travelled . I know somewhere I am doing wrong. Please help me solving this problem.
  2. jcsd
  3. Apr 14, 2009 #2
    it appears this problem could be split into different sections, try to find out what you can about the bus in each of these sections, then see where that takes you
  4. Apr 14, 2009 #3
    also do you have the answers so we can see what we are aiming for?
    i would also check the time you got for substituting u=0 and a=1.5, i didnt get t=9.
  5. Apr 14, 2009 #4
    The time taken to reach 9m/s with a= 1.5m/s^2 is v=u+at => 9= 0+1.5t => t ( initial)= 6sec.
    the initial distance travelled to reach from 0 to 9m/s with a =1.5 and t =6 is s= ut+i/2 at^2

    => the distance s(initial)= 0+1/2 * 1.5 * 6*6 = 27.

    The time taken to stop from u= 9m/s with a = -2m/s^2, v =0 is v= u+at => t (final)= 4.5sec.

    The final distance travelled when the bus was decelerating before stop with v=0,
    u=9m/s, a -2m/s^2 is v^2 = u^2 +2as => s( final)= 20.25m

    So the distance travelled with 9m/s continuously= total distance- s(initial)-s(final)
    => 400-27-20.25= 352.75m

    the time taken to travel 352.75 m with the velocity 9m/s is t= s/v = 352.75/9= 39.19 sec

    so the total time = t (initial)+ t ( final) + t = 6+4.5+39.19 = 49.69 sec

    But in the text book the answer was given as 50 sec.

    Please let me know where I have done wrong
  6. Apr 14, 2009 #5
    yeah i got the same asnwer!
    well the text book has obviously rounded the answer to the nearest second thats all!
    well done =]
  7. Apr 14, 2009 #6
    Thank you RoryP
  8. Apr 14, 2009 #7
    Looks like you already solved this, good job. Using different words to do the same thing:


    So you found t1 = 6
    And you found that t3 = 4.5

    Instead of d = vt + .5at^2 to find d1 and d3, you can also use average velocity. Vave= (v + v')/2 .

    So Vave1 = 4.5 m/s , Vave2 = 9 m/s , and Vave 3 is 4.5 m/s .

    d = (Vave1)(time1) + Vave2(time2) + Vave3(time3)

    You can solve for time2 and add it with t1 and t3 from above to get 49.69 s.
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