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zaphod

  • 2 years ago

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?

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  1. zaphod
    • 2 years ago
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    @Callisto @davian @inkyvoyd

  2. Callisto
    • 2 years ago
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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 :(

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

  4. Callisto
    • 2 years ago
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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?

  5. zaphod
    • 2 years ago
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    no the books answer is tht, i tried nd i dnt get it o.O

  6. Callisto
    • 2 years ago
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    What do you get for (2) - (1) ???

  7. Callisto
    • 2 years ago
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    I meant my way to get the answer doesn't make sense..

  8. zaphod
    • 2 years ago
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    so theres no solution?

  9. Callisto
    • 2 years ago
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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?

  10. zaphod
    • 2 years ago
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    yes i agree:)

  11. Callisto
    • 2 years ago
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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.

  12. Callisto
    • 2 years ago
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    I know it doesn't make sense at all...

  13. zaphod
    • 2 years ago
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    it does:) \[\sqrt{8}--->2\sqrt{2}\]

  14. Callisto
    • 2 years ago
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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. :(

  15. zaphod
    • 2 years ago
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    can u explain ur assumption?

  16. zaphod
    • 2 years ago
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    i understand b^2 - 4ac = 0

  17. zaphod
    • 2 years ago
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    thanks alot :)

  18. Callisto
    • 2 years ago
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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

  19. zaphod
    • 2 years ago
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    yes..evn i thought the same thng

  20. Callisto
    • 2 years ago
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    But there is a bug? :|

  21. zaphod
    • 2 years ago
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    wt u mean?

  22. Callisto
    • 2 years ago
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    Hmm.. I don't know :|

  23. zaphod
    • 2 years ago
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    anyway thnks once again :)

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

  25. ash2326
    • 2 years ago
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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 \[b^2-4ac \ge0\]

  26. Xishem
    • 2 years ago
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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.

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