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zaphod Group Title

A bus moves away from rest at a bus stop with an acceleration of 1ms^-2. As the bus stars to move a man who is 4m behind the stop runs with a constant speed after the bus. If he just manages to catch the bus find his speed?

  • one year ago
  • one year ago

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  1. zaphod Group Title
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    @Callisto @davian @inkyvoyd

    • one year ago
  2. Callisto Group Title
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    I don't know if it works. |dw:1346211401321:dw| Let s be the distance travelled by bus, distance travelled by the man = s+4 Time for the bus and man travelled is the same For the bus: s = ut + (1/2) at^2, where u = 0ms^-1, a = 1ms^-2 For the man, s + 4 = vt No... it doesn't work. I need one more equation :(

    • one year ago
  3. zaphod Group Title
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    yup..i tried to buy the answer is \[2\sqrt{2} ms^{-1}\]

    • one year ago
  4. Callisto Group Title
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    I know it doesn't make sense... s = ut + (1/2) at^2, where u = 0ms^-1, a = 1ms^-2 So, s = 0 + (1/2)(1)(t^2) s = t^2 / 2 -(1) s + 4 = vt -(2) (2)- (1), what do you get?

    • one year ago
  5. zaphod Group Title
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    no the books answer is tht, i tried nd i dnt get it o.O

    • one year ago
  6. Callisto Group Title
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    What do you get for (2) - (1) ???

    • one year ago
  7. Callisto Group Title
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    I meant my way to get the answer doesn't make sense..

    • one year ago
  8. zaphod Group Title
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    so theres no solution?

    • one year ago
  9. Callisto Group Title
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    No no no,... You don't even answer my question :| s = t^2 / 2 -(1) s + 4 = vt -(2) (2) - (1): 4 = vt - t^2/2 Agree?

    • one year ago
  10. zaphod Group Title
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    yes i agree:)

    • one year ago
  11. Callisto Group Title
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    Now, rearrange it. -t^2/2 + vt - 4 = 0 Quadratic equation. I take an assumption here: there is only one intercept for the x-axis, which means the man can only catch the bus at one time one intercept: delta = 0 That is v^2 - 4(-1/2)(-4) = 0 Solve v.

    • one year ago
  12. Callisto Group Title
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    I know it doesn't make sense at all...

    • one year ago
  13. zaphod Group Title
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    it does:) \[\sqrt{8}--->2\sqrt{2}\]

    • one year ago
  14. Callisto Group Title
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    I don't think it is the normal way to work out the solution... That's why I said this method doesn't make sense. :(

    • one year ago
  15. zaphod Group Title
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    can u explain ur assumption?

    • one year ago
  16. zaphod Group Title
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    i understand b^2 - 4ac = 0

    • one year ago
  17. zaphod Group Title
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    thanks alot :)

    • one year ago
  18. Callisto Group Title
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    That's the most nonsense point. Since the bus is accelerating, the man doesn't run faster to catch the bus, when the man misses that chance, the bus would be further away from him and it's not possible to catch the bus if he doesn't accelerate

    • one year ago
  19. zaphod Group Title
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    yes..evn i thought the same thng

    • one year ago
  20. Callisto Group Title
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    But there is a bug? :|

    • one year ago
  21. zaphod Group Title
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    wt u mean?

    • one year ago
  22. Callisto Group Title
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    Hmm.. I don't know :|

    • one year ago
  23. zaphod Group Title
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    anyway thnks once again :)

    • one year ago
  24. Callisto Group Title
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    Welcome. but better wait if some others can give a more reasonable way to do this question! I'm sorry!

    • one year ago
  25. ash2326 Group Title
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    Let the bus covers a distance x, while the man runs so the man has to run for a distance of (4+x) meters Let this time be t and the constant speed of man be c \[(4+x)=c\times t\] now for the bus \[S=ut+\frac 12 at^2\] \[x=0+\frac 12 1 t^2\] so \[t^2=2x\] we have the other equation as \[4+x=ct\] x=t^2/2 so we have \[8+t^2=2ct\] rearranging we get \[t^2-2ct+8=0\] for a solution to exist \[b^2-4ac \ge0\]

    • one year ago
  26. Xishem Group Title
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    This defines the movement of the bus:\[x_b=x_{0b}+v_{0b}t+\frac{1}{2}a_bt^2\]And this defines the movement of the man:\[x_m=x_{0m}+v_{0m}t+\frac{1}{2}a_mt^2\]We want to see what the velocity of the man is when it's x position is the same as the bus', so let's set their positions equal to eachother:\[x_{0b}+v_{0b}t+\frac{1}{2}a_bt^2=x_{0m}+v_{0m}t+\frac{1}{2}a_mt^2\]We can express the bus' initial position in terms of the man's initial position:\[x_{0b}=x_{0m}+4m\]We also know that the acceleration of the man is 0 (constant speed) and that the initial velocity of the bus is 0. This means we can cancel some things out in our equation:\[(x_0m+4m)+\frac{1}{2}a_bt^2=x_0m+v_{0m}t\]x0 of the man cancels:\[4m+\frac{1}{2}a_bt^2=v_{0m}t\]Now we are down to 2 unknown variables. We need to express one of the two unknowns (v_0m and t) in terms of eachother, so let's look at another equation:\[v_{fm}=v_{0m}+a_mt \rightarrow v_{fm}=v_{0m}\]We know that if the man JUST catches the bus, their velocities will be equal at the time that the man catches it, so we can assume that:\[v_{fm}=v_{fb}\]The final velocity of the bus can be defined as:\[v_{fb}=v_{0b}+a_bt \rightarrow v_{fb}=a_bt\]Since final velocities are equal:\[v_{fb}=a_bt \rightarrow v_{fm}=a_bt\]Let's plug this back into our equation:\[4m+\frac{1}{2}a_bt^2=v_{0m}t \rightarrow 4m+\frac{1}{2}a_bt^2=(a_bt)t\]\[4m+\frac{1}{2}a_bt^2=a_bt^2\]\[4m=\frac{1}{2}a_bt^2 \rightarrow t=\sqrt{\frac{8m}{a_b}}\]\[t=\sqrt{\frac{8m}{+1\frac{m}{s^2}}}=\sqrt{8}s\]We can now plug this back into one of our old position equations to find v_0m (and therefore v_fm):\[4m+\frac{1}{2}a_bt^2=v_{0m}t \rightarrow v_{0m}=\frac{4m+\frac{1}{2}a_bt^2}{t}=\frac{4m}{t}+\frac{1}{2}a_bt\]\[v_{0m}=v_{fm}=\frac{4m}{t}+\frac{1}{2}a_bt \rightarrow v_{fm}=\frac{4m}{\sqrt{8}s}+\frac{1}{2}(1\frac{m}{s^2})(\sqrt{8}s)\]\[v_{fm}=+2.83\frac{m}{s}\]Which matches up with other answers given.

    • one year ago
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