[Kinematics] Question involving quadratic equation.

[Hn.]

An object accelerates at the rate of 1.30 m/s^2 over a displacement of 11.8m, reaching a final velocity of 7.00 m/s. For what length of time was it accelerating? Explain the answers.

NOTE: Actually, when I first was about to post this, I finally had gotten the answer I needed. However, I cannot explain why I got these two answers (So now a new question.

a = 1.30 m/s^2 , d = 11.8m , v = 7 m/s , t = ?

d = v2t - 1/2at^2

11.8 = 7t - 0.65t^2

0 = 0.65t^2 - 7t + 11.8

-b \pm\sqrt{b^2-4ac}
/2a (The entire equation above)

7 \pm\sqrt{-7^2-4(0.65)(11.8)}
/2(0.65) (The entire equation above)

= 7 \pm\sqrt{49-30.68}
/1.3 (The entire equation above)

Answer:

(7 + 4.28)/2 = 8.68 s

or

(7 - 4.28)/2 = 2.09 sNow I know the answer but, I can't explain why there could be two answers. Can anyone explain?

 

[LowlyPion]

Welcome to PF.

I don't think your equation is correct.

X = X_o +V_o*t + 1/2*a*t²

Displacement is determined off the initial conditions, not the final.

In your construction you have used what you call final velocity as initial, implying a question that would be how long would it take a ball rolled up an incline at 7m/s to go 11.8m if it is decelerating at 1.3m/s²?

If your 7 m/s is your final velocity, then it implies an initial velocity?

V_f² = V_i² + 2*a*x

49 = V_i² + 2*1.3*11.8

V_i² = 18.32

V_i = ± 4.28m/sec

Now you can solve for time with the first equation knowing what the initial V is.

 

[BAnders1]

The reason you receive two answers is due to the fact that the initial velocity may be in the opposite direction of the acceleration, thus the particle will have to turn around before accelerating to its final velocity of 7.00 m/s.

For example, consider throwing a ball downward from 10m high. The ball will accelerate with gravity and reach the ground ( a displacement of 10m). Now consider throwing the ball straight upward with the same initial velocity. The ball will reach some maximum height, then it will stop, turn around, and fall to the ground. Upon hitting the ground it will have the same velocity as ball 1, the same constant acceleration, and the same final displacement. The only difference is that ball 2 was initial thrown in the opposite direction.

 

[Hn.]

LowlyPion, you are correct that 7 m/s is a final velocity. However, the equation I used did not require me to find out the Initial Velocity to find the time.

BAnders1, thanks for the information.

 

[LowlyPion]

LowlyPion, you are correct that 7 m/s is a final velocity. However, the equation I used did not require me to find out the Initial Velocity to find the time.

I see so you solved it backwards, assuming it was slowing from 7m/s, which works as well, though for me anyway, I think it makes understanding your 2 time solutions more challenging.

 

[Hn.]

Answer:

(7 + 4.28)/2 = 8.68 s

or

(7 - 4.28)/2 = 2.09 s


Since you stated the initial velocity was 4.28, I'm guessing that:

t = (7 \pm 4.28)/2

= (v1 \pm v2)/2

So it is equal to say that:

t = 1/2(v1\pmv2)

= 1/2(7\pm4.28)

The initial velocity is found along the way when we do the quadratic equation (that is what I'm assuming because of the 4.28 appearing in the quadratic equation and as the resultant initial velocity; but then again it could be mere coincidence).


EDIT: I realize that your equation is also correct:

Vf2 = Vi2 + 2*a*x

49 = Vi2 + 2*1.3*11.8

Vi2 = 18.32

Vi = ± 4.28m/sec

Is equal to using part of the quadratic equation:

Vi = \sqrt{-b^2-4(ac)}

= \sqrt{-7^2-4(0.65)(11.8)}

= \sqrt{49-30.68}

= \sqrt{18.32} = 18.32^2

= 4.28 m/s

Then all I had to do was do t = 1/2(Vi±Vii).


Thanks for all the help, I now understand it a lot more than before.

 

[LowlyPion]

Equating time to differences in velocities is a dangerous game.

To be more precise using the result that

Vi = ± 4.28m/sec

And plugging into a new equation for displacement yields:

11.8 = 4.28*t + 1/2*a*t²

.65t² + 4.28*t - 11.8 = 0

yields 2 results : 2.09 sec and -8.68 sec
Using the other velocity -4.28 yields 8.68 sec and -2.09sec

These answers then are both consequences of the difference in direction of the initial velocity.