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Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 decade ago

what am i doing wrong in this math/physics question?

i am trying to figure out a physics question and i cant seem to get this new formula to work for me.

The question says:

Assume there is a car moving to the left at an initial speed of 14 m/sec. In addition, assume the car is experiencing a constant acceleration of 5.3 (m/sec)/sec to the right. What will its velocity be at 1.7 seconds and at 3 seconds?

And im almost 100% sure that this formula is used:

V=Vo+at

I cant make it look like how it is written but the o after the v is suppose to be smaller. I am new at learning this formula and in some cases it seems to be that i can times +at or divide it i dont know how to know which to do times or divide how can i tell?

so anyway:

V=Vo+at (Kinematic) or (volocity formula)

numbers are:

-14 m/sec

+5.3 (m.sec)/sec

1.7 time and 3 time

I am really stumped with this and i cant move on in my learning untill i can figure it out.

6 Answers

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  • 1 decade ago
    Favorite Answer

    that is the correct formula, yes.

    Vo is the initial velocity, ie -14 m/s.

    a is the acceleration, ie 5.3 m/s/s

    t is time, either 1.7 or 3 s

    at 1.7 sec, velocity will be:

    -14 + (5.3 x 1.7) = - 4.99

    so the car will be moving to the left at 4.99 m/s

    at 3 sec velocity will be:

    -14 + (5.3 x 3) = 1.9

    so the car will be moving to the right at 1.9 m/s

  • 1 decade ago

    The o after the V is "not" meaning original or initial. Combined with the V means initial velocity.

    Vo=14m/s

    a=5.3m/s²

    you are given 2 t's to find the velocity at they are:

    t=1.7s and 3s

    Yes you use V=Vo+at

    assuming left is negative and right is positive

    V=-14+(5.3)(1.7)

    =-5m/s

    V=-14+(5.3)(3)

    =2m/s

    I don't see how you can get stumped. All you have to do is plug in your numbers and use a calculator to calculate V, the final speed.

  • 1 decade ago

    Okay. Let's say that left = negative. Number line logic works okay here.

    So we've got the following information:

    u = 14ms^-1 <-- initial velocity

    a = -5.3ms^-2

    t = two variables.

    v = ? <-- final velocity

    I think we used different symbols to you. Oh well. It works the same. :)

    So we use the formula v = u + at, which is the same one you've got.

    Right, they've given us two variables for time. So it's a simple formula-plugging question.

    @ t = 1.7s

    v = 14 + (-5.3)(1.7)

    v = 14 + (-9.01)

    v = 4.99ms^-1

    @ t = 3s

    v = 14 + (-5.3)(3)

    v = 14 + (-15.9)

    v = -1.9ms^-1

    This means that at some point between t = 1.7s and t = 3s, the car began travelling in reverse.

    Did this help? :\

  • 1 decade ago

    You are on the right track. In fact you are so ‘on the right track’ that it is hard to tell what you are having a problem with. Perhaps it is simply a matter of confidence!

    Here is an easy way to solve problems like this in a way that you can be sure that you are doing it correctly.

    Divide you page into three columns. In the first column put the information you have been given in the problem. Be sure to include the information you are looking for as well.

    In the middle column put the equation you are going to use and if necessary solve the equation for the unknown from the first column.

    In the third column do the math work required by the equation in the second column. Be sure to do the math with both the numbers and with the units that go with the numbers.

    So, your problem looks like this:

    vi = 14m/s vf = vi + at vf1 = 14m/s + (-5.3m/s/s)(1.7s)

    a = -5.3m/s/s vf1 = 14m/s + (-9.0m/s)

    vf1 = ?m/s vf1 = 5m/s

    vf2 = ?m/s vf2 = 14m/s + (-5.3m/s/s)(3s)

    t1 = 1.7s vf2 = 14m/s + (-15.9m/s)

    t2 = 3s vf2 = -1.9m/s

    (In the answer box the setup of this problem looks great; but in the actual answer box it looks terrible! You may find it helpful to copy the solution on your own paper. This is the information in the first column:

    vi = 14m/s

    vf1 = ?m/s

    vf2= ?m/s

    a = -5.3m/s

    t1 = 1.7s

    t2 = 3s

    The second column looks like this: vf = vi + at

    The third column looks like this:

    vf1 = 14m/s + (-5.3m/s/s)(1.7s)

    vf1 = 14m/s + (-9.0m/s)

    vf1 = 5m/s

    vf2 = 14m/s + (-5.3m/s/s)(3s)

    vf2 = 14m/s + (-15.9m/s)

    vf2 = -1.9m/s

    In the first answer you can see that the direction of the motion has not changed and the object is still going to the left. In the second answer the negative indicates that the direction of the object has changed and now the object is traveling to the right.

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  • 1 decade ago

    You are right in using the formula

    V = -14 + 5.3x1.7 = -4.99 m/s , negative sign shows the vel. is to the left

    V = -14 + 5.3 x 3 = 1.9 m/s , Now the car is moving towards the right . The velocity has changed sign.

  • Anonymous
    1 decade ago

    V is your velocity at the given time.

    Vo is your original velocity

    A is the acceleration

    T is the time

    V = -14m/s + 5.3*1.7

    V = -4.99m/s @ 1.7 seconds

    V = -14 m/s + 5.3 * 3

    V = 1.9 m/s @ 3 seconds

    This means that at 1.7 seconds, the object is still moving to the left. At three seconds, it it now slowly moving to the right.

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