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Total time travelled

  • Thread starter phynaive
  • Start date
  • #1
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Homework Statement


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?


Homework Equations



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

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.
 

Answers and Replies

  • #2
76
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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
 
  • #3
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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.
 
  • #4
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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
 
  • #5
76
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yeah i got the same asnwer!
well the text book has obviously rounded the answer to the nearest second thats all!
well done =]
 
  • #6
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Thank you RoryP
 
  • #7
18
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Looks like you already solved this, good job. Using different words to do the same thing:

|---1----->|---2--------->|---3------->|

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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