Combination Kinematics Problem - Quadratic Formula

**shadow6** · March 16th 2010, 03:20 PM

About 3 weeks ago I posted here because I needed help for a bonus question. Though I was told that I would need to do this myself due to it being a bonus. The test for it passed and I obviously didn't get it. However, I'm curious as to how to solve it. I was told by a friend that the quadratic formula is involved in this. The thing is, I'm not exactly sure how to apply it to this question:

"A boy and a ball are at rest at the top of an incline. The ball rolls down the incline with an acceleration of 0.2m/s^2. Twelve seconds later the boy chases after the ball with an acceleration of 9.0m/s^2. The length the incline is 60m. Draw a diagram before solving the problem.

a. Will the boy catch the ball before the end of the incline? Hint "When are their displacements equal?"

What am I supposed to solve for? Is it the velocities of the boy and the ball?

b. What is the velocity of the boy when he catches the ball.

c. At what time does the boy catch the ball?"

I'm guessing that the kinematic equation $d = (v_1)(t) + \frac{1}{2}a(t)^2$ will be involved. I think that I will have to rearrange this equation into the form $ax^2 + bx + c$ and then solve using the quadratic formula. Though, what equations will I use to represent the displacement of the boy and the ball?

**BabyMilo** · March 17th 2010, 01:34 AM

Originally Posted by shadow6

About 3 weeks ago I posted here because I needed help for a bonus question. Though I was told that I would need to do this myself due to it being a bonus. The test for it passed and I obviously didn't get it. However, I'm curious as to how to solve it. I was told by a friend that the quadratic formula is involved in this. The thing is, I'm not exactly sure how to apply it to this question:

"A boy and a ball are at rest at the top of an incline. The ball rolls down the incline with an acceleration of 0.2m/s^2. Twelve seconds later the boy chases after the ball with an acceleration of 9.0m/s^2. The length the incline is 60m. Draw a diagram before solving the problem.

a. Will the boy catch the ball before the end of the incline? Hint "When are their displacements equal?"

What am I supposed to solve for? Is it the velocities of the boy and the ball?

the ball rolls for 12s,

then using x= ut+0.5at^2

x=0t+0.5*0.2*12^2= 14.4m

v=u+at=0+0.2*12= 2.4ms^-1

after 20s,

the equation for the ball is x=2.4t+0.5*0.2*t^2

and for the boy the equation is x=0t+0.5*0.9*t^2.

x=x so 2.4t+0.5*0.2*t^2=0t+0.5*0.9*t^2

rearrange to find t.

sub t into the equation for the ball is x=2.4t+0.5*0.2*t^2.

then add that to 14.4m = total distance travelled by the ball.

so if that exceeds 60m, if not, you have your answer.

Hope this helps!

**shadow6** · March 17th 2010, 06:18 AM

Thanks very much for the reply!

There was a slight mistake I made in copying the question here though. The acceleration was supposed to be 0.9 not 9.0. However, I was able to correct that in my calculations.

After making the equations for the boy and the ball equal, I simplified them into $-0.35m/s^2(t^2) + 4m/s(t) + 0$ then plugged that into the quadratic formula. My $t$ values were $0s$ and $11.43s$ . Next, I subbed that into the equation of the ball after 20s and got $58.79m$ . Finally, I added that to $40m$ and got $98.79m$ . Does this look about right? The final answer is larger than I had anticipated.

**shadow6** · March 17th 2010, 06:21 AM

Basically, my answer is saying that the ball and the boy had an equal displacement of 98.79m after 11.43s. However, this is before the boy even had a chance to go down the ramp (before 12s), so how could that be?

**Haytham** · March 17th 2010, 07:41 AM

after 20s,

the equation for the ball is x=4t+0.5*0.2*t^2

and for the boy the equation is x=0t+0.5*9*t^2.

there is 2 mistakes in this

1st one is the 20s it should be 12 according with the text, that changes the 4m/s value in the equation x=4t+0.5*0.2*t^2

the 2nd is also in the ball equation x=x0 + vo* t + 0.5* a* t²
the xo value isnt zero as the ball was rolling down for a while after the boy starts to chase it

you need to find it like babymilo did it to find the value 40m but you need to correct the 20s value to 12.

after correcting the ball equation you can solve the quadratic equation

**BabyMilo** · March 17th 2010, 08:15 AM

Originally Posted by Haytham

there is 2 mistakes in this

1st one is the 20s it should be 12 according with the text, that changes the 4m/s value in the equation x=4t+0.5*0.2*t^2

the 2nd is also in the ball equation x=x0 + vo* t + 0.5* a* t²
the xo value isnt zero as the ball was rolling down for a while after the boy starts to chase it

you need to find it like babymilo did it to find the value 40m but you need to correct the 20s value to 12.

after correcting the ball equation you can solve the quadratic equation

i dont understand what you are saying about the 2nd mistake.

and thanks, will change the 1st mistake now.

**shadow6** · March 17th 2010, 09:51 AM

Alright so I simplified the equations to $0 = 0.35m/s^2(t)^2 + 2.4m/s(t) + 14.4m$ . When I plug this into the quadratic formula, I keep ending up with a negative square root:

$<br /> t = \frac{2.4m/s \pm \sqrt{(-2.4m/s)^2 - 4(0.35m/s^2)(-14.4m)}}{2(-0.35m/s^2)}$

$<br /> t = \frac{2.4m/s \pm \sqrt{(5.76m^2/s^2) - (20.16m^2/s^2)}}{(0.7m/s^2)}$

$<br /> t = \frac{2.4m/s \pm \sqrt{(-14.4m^2/s^2)}}{(0.7m/s^2)}$

I noticed that the negative square root I end up with in this formula is the same (excluding the negative and units of measurement) value that I got for the distance of the ball after it rolls for 20s (I got 14.4m after the correction). So how do I solve this now, if the values that I plugged into the equation are correct? I'm pretty sure imaginary numbers aren't involved in this, as it wont be another year until I even get to that topic... But then when I plug the $a$ , $b$ , and $c$ ( $0.35$ , $2.4$ , and $14.4$ ) values into a calculator on Quadratic Equation Calculator , my $x_1$ value is $-3.4286 + 5.421 i$ and my $x_2$ value is $-3.4286 - 5.421 i$ . Was the simplifying done correctly?

**skeeter** · March 17th 2010, 10:40 AM

"A boy and a ball are at rest at the top of an incline. The ball rolls down the incline with an acceleration of 0.2 m/s^2. Twelve seconds later the boy chases after the ball with an acceleration of 0.9 m/s^2. The length the incline is 60m. Draw a diagram before solving the problem.

a. Will the boy catch the ball before the end of the incline? Hint "When are their displacements equal?"

b. What is the velocity of the boy when he catches the ball.

c. At what time does the boy catch the ball?"

... the correct acceleration of the boy has been noted.

in 12 seconds, the ball's displacement is $\frac{1}{2}(0.2)(12^2) = 14.4$ m. the ball's speed after 12s is $(0.2)(12) = 2.4$ m/s

let $t = 0$ be the time the boy starts after the ball.

position of the ball relative to the top of the incline ...

$x = 14.4 + 2.4t + (0.1)t^2$

position of the boy relative to the top of the incline ...

$x = (0.45)t^2$

setting the two positions equal ...

$(0.45)t^2 = 14.4 + 2.4t + (0.1)t^2$

$(0.35)t^2 - 2.4t - 14.4 = 0$

multiply every term by 20 to clear the decimals ...

$7t^2 - 48t - 288 = 0$

$t = \frac{48 \pm \sqrt{48^2 - 4(7)(-288)}}{14}$

positive solution ... $t \approx 10.7$ seconds

after that time, the ball and boy are at position $x = (0.45)t^2 \approx 51.5$ m relative to the top of the incline.

boy's speed when he catches the ball ...

$v = at = (0.9)t \approx 9.63$ m/s

**shadow6** · March 17th 2010, 11:39 AM

TYVM all for the help.

Just have one quick question. The 10.7s value that I got... that's the time it takes for the boy to catch up to the ball, right? I mean, it wouldn't make sense if that was the time it took for the boy to catch the ball immediately after the ball was released.

**skeeter** · March 17th 2010, 11:53 AM

Originally Posted by shadow6

TYVM all for the help.

Just have one quick question. The 10.7s value that I got... that's the time it takes for the boy to catch up to the ball, right? I mean, it wouldn't make sense if that was the time it took for the boy to catch the ball immediately after the ball was released.

the 10.7 s value is 22.7 s after the ball is released.

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