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? asked in Science & MathematicsPhysics · 2 months ago

Physics help. I know the answers but why are they the correct answer.?

9) A bullet fired straight upwards experiences an acceleration due to air resistance of 1 m/s2 during its entire flight (this is an approximation; in reality it would depend on velocity.) If it leaves the barrel of the gun at 300m/s , with what speed does it hit the ground?

- I know the answer is 270 m/s, but why? What equation was used?

10) An object with an initial velocity of 5m/s slows to 1m/s over a distance of 10m. What is the magnitude of its average acceleration?

- So, the answer is 1.2 m/s^2, but when I solved the problem, I got -1.2m/s^2. Why isn't it negative since it's decreasing in speed?

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  • 2 months ago
    Favourite answer

    You can do this quite simply using the standard equations of motion (which is what I assume you are supposed to do).

    "Acceleration" can be positive or negative, depending upon which direction you choose as your reference.  You have to be careful to keep the signs consistent with your chosen reference when you write your equations.  That is where your problem lies in number 10.

    You have an initial speed, a final  speed, and a distance.  This calls for the equation  v^2 = u^2 + 2as.

    Now, entering those numbers directly into that equation gives you

    1^2 = 5^2 + 2(a)10

    so a = (1 - 25) / 20 = -1.2 m/s

    But the positive sign in the equation assumes the object has (positive) acceleration in the direction of its motion.  But it is slowing down, so its acceleration is NEGATIVE, and you should have written

    1^2 = 5^2 + 2(-a)10

    which gives a value of +1.2m/s which is measured BACKWARDS, because it is slowing down.

    So choose your directions first, and make sure that your equations match.  And remember, depending upon that, acceleration can come out either positive or negative.

    ___________________________________________

    A similar consideration enters into the question 9.

    The first thing to find out is how high does the bullet reach?

    Using the same equation as above, you have u = 300 m/s, v = 0, you want to find the height (s), but acceleration - the tricky bit.  Gravity always acts downwards (or usually anyway, so the positive direction is usually taken as down).

    Now, what IS the acceleration? Yes, gravity at -9.81 m/s, huh (negative because the bullet is going UP, opposite to the direction of gravity)?   BUT  air resistance (therefore slowing the bullet down MORE) at 1 m/s ADDS to the resistance due to gravity, so -9.81 becomes -10.81m/s, hence the equation is

    0^2 = 300^2 - 2(10.81)s, therefore s = 300^2 / 21.62 = 4162.81 m.

    Now repeat for the fall of the bullet.  But this time, acceleration is POSITIVE, and it is DECREASED by 1 m/s because air RESISTANCE is always negative.  Therefore a = (9.81 - 1) = 8.81 m/s

    v^2 = 0^2 + 2(8.81)(4162.81) . . . . (note positive sign)

    which should give you the answer you want.

  • ?
    Lv 7
    1 month ago

    9

    hmax = Vo^2/2(g+1) = 300^2/21.6 = 4167 m

    (0-4167) = (-g+1)/2*t^2

    -8333 = -8.8*t^2

    t = √-8333/-8.8 = 30.8 sec

    V = (-9.8+1)*t = -8.8*30.8 = -271 m/sec

    10

    Vf^2-Vi^2 = 2*a*d  

    2*10*a = 1-25

    a = -24/20 = -1.2 m/sec^2 

  • 2 months ago

    9)

    Vy(t) = Vyo - 10.8t = 0 at t = Vyo/10.8Vyo² = 2*10.8*H = 300² so H = 12,500/3Vyf² = 2*8.8*12,500/3 = 271² so Vyf is about 270m/s with 2 significant figurs.

    10) Use Vf² - Vi² = 2*a*d again.

    1² - 5² = 2*a*10

    -24/20 = -12/10 = -1.2m/s²  It is deceleration so technically it is negative. Magnitude is what is wanted which is absolute value.

  • 2 months ago

    assuming that is "de-acceleration due to air resistance" as air resistance cannot cause acceleration.

    going up, acc = 9.8 – 1 = 8.8 m/s² (neg)

         air resistance slows it down

    going down, acc is still the same 9.8 – 1 = 8.8 m/s²

         air resistance slows it down again

    so it would hit the ground at the speed it left, 300 m/s

    book is wrong.

    v² = v₀² + 2ad

    1² = 5² + 2a10

    a = –24/20 = –1.2 m/s²

    you are correct. book is wrong

    Equations of motion (straight line, constant acc)

    d = ½at² + v₀t + d₀

        d is displacenemt

        v₀ is initial velocity

        d₀ is initial position

    v = v₀ + at

    v² = v₀² + 2ad

    F = ma

    a = Δv/Δt

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  • 2 months ago

    Problem 9.

    You need to use conservation of energy and the work-energy theorem.  THe initial energy of the bullet, Ei is

    Ei = 1/2 mv0^2  where v0 = 300 m/s

    Now the bullet eventually reaches a height h determined by the initial energy minus the work done by air resistance.  The work done is

    W = F*h = ma*h  where a = 1 m/s^2 and m = mass of bullet

    so

    mgh = 1/2 mv0^2 - mah  and  solve for h

    (g+a)h = 1/2 v0^2  -->  h = v0^2/(2(g + a))

    Now the return trip - the air resistance still does work to slow the bullet down so the final speed vf can be found using conservation of energy

    mgh - mah = 1/2 mvf^2

    (g -a)*v0^2/(2*(g+a)) = 1/2 vf^2

    vf = v0*sqrt{(g -a)/((g+a)) } = 270.8 m/s

    In problem 10 they asked for the MAGNITUDE of the decreasing acceleration.  That means the want you to give the absolute value ofr the acceleration, i.e. a positivie number

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