# 0 = t(7.8 x 10² m/s - 4.9 m/s² t)

1. Nov 5, 2007

### viet_jon

[SOLVED] 0 = t(7.8 x 10² m/s - 4.9 m/s² t)

1. The problem statement, all variables and given/known data

0 = ( 7.8 x 10² m/s ) x t + ½ x (-9.5 m/s²) x t²

I have time, represented by t in the equation twice....how would I figure out what time is? This is probably simple math, but being out of school for 7 years now, someone please refresh my memory.

2. Relevant equations

0 = t(7.8 x 10² m/s - 4.9 m/s² t)
0 = t

or 0 = (7.8 x 10² m/s - 4.9 m/s² t)

4.9 m/s² t = 7.8 x 10² m/s

t = (7.8x10²m/s) / (4.9 m/s²)
= 1.6 x 10² s

3. The attempt at a solution

2. Nov 5, 2007

### Sourabh N

I didn't understand you, you already have the two values for time. These represent the two times at which your condition is met.

3. Nov 5, 2007

### S[e^x]=f(u)^n

you have your two times.... they're written right there

Last edited: Nov 5, 2007
4. Nov 5, 2007

### viet_jon

opps...sorry guys....forgot to mention, the answer shown is not my answer, it's the answer in the back of the book.

I don't understand how they got it. I only know how to isolate 't' (time), if there's one variable. i.e. 1 = t + 2.....isolated would be 1 - 2 = t.

In the original equation, there's 't' twice......how would I isolate that?

5. Nov 5, 2007

### Feldoh

If we leave the equation in a symbolic form and factor out a t we get:

0 = t(v+0.5gt)

When does the RHS = 0?

ax^2+bx=0 you could consider the equation like that if it helps...

6. Nov 5, 2007

### CompuChip

Let's drop all the units for a while, then your equation is of the type
$$0 = a t^2 + b t$$
where a, b are just numbers and you want to solve for t.

You can use the "no-thinking-involved" way and just plug it into the abc-formula, which says that the solutions for
$$0 = a t^2 + b t + c$$
in general are
$$t = \frac{ - b \pm \sqrt{b^2 - 4 a c} }{2 a}$$
(if you choose the plus or the minus you get two solutions).

But in this case there is an easier way, and that's noticing that you can bring a t outside the brackets:
$$0 = (a t + b)t,$$
if you work out the brackets you'll see that this is the same as the original equation.
Now you have a product of two things, namely t and a t + b. When is a product of two numbers zero?

7. Nov 5, 2007

### viet_jon

now I'm more confused....what's v? and what's gt?

8. Nov 5, 2007

### Feldoh

Your equation 0 = ( 7.8 x 10² m/s ) x t + ½ x (-9.5 m/s²) x t²

was derived from $$x_f =x_0+v_0t+\frac{1}{2}gt^2$$

v and g are just constants (7.8 x 10² m/s for v and -9.5 m/s² for g)

Last edited: Nov 5, 2007
9. Nov 5, 2007

### viet_jon

where does c come from?

and how does 0= at2 + bt + c

work out to be

$$t = \frac{ - b \pm \sqrt{b^2 - 4 a c} }{2 a}$$

???

i've never learned this math before, is there a name for this? I tried googling, but don't know what this is called.

10. Nov 5, 2007

### Feldoh

It's called the quadratic formula -- basically it's used to solve quadratic equations (equations that have a variable to the second power t^2)

What's that highest level of math you've taken?

11. Nov 5, 2007

### viet_jon

grade 12 ..... but that's 7 years ago.

been dicking around ever since, forgot everything.

i'm taking Uni prep Physics grade 12 right now.

12. Nov 5, 2007

### Feldoh

Ah ok... Well there's an easier way then using that formula:
0=(7.8 x 10² m/s) x t + ½ x (-9.5 m/s²) x t²

Would you agree that we can factor out a "t" and get:

0=t((7.8 x 10² m/s) + ½ x (-9.5 m/s²) x t)

Would you also agree the anything multiplied by 0 will equal 0?

So if t=0 we get:

0=0((7.8 x 10² m/s) + ½ x (-9.5 m/s²) x 0)
0=0(7.8 x 10² m/s + 0)
0=0

So we know that t=0 is a solution. But that's not the only solution.

Would you agree that 0=t((7.8 x 10² m/s) + ½ x (-9.5 m/s²) x t) when

(7.8 x 10² m/s) + ½ x (-9.5 m/s²) x t = 0

since that would give us 0 = t x 0 which is 0

13. Nov 5, 2007

### CompuChip

Ah, since you were asking a physics-based question, I assumed a high-school level of mathematics. So I guess you have some reading up to do (as a side note, if you want to really learn algebra thoroughly, solving quadratic equations may not be the easiest way to start. )

14. Nov 5, 2007

### viet_jon

my heads spinning on the physics alone. I'm gonna try to figure out quadratic equations before I attempt other methods.

15. Nov 5, 2007

### viet_jon

before this lesson, I ran into something I didn't know.....trigonometry. Had to download ebooks to figure it all out. I got most the basics down now.

16. Nov 5, 2007

### viet_jon

I'd like to comment on how wicked this forum is. Thnkx guys/.....I really appreciate all the help. I hope this forum stays as active as it is now.

cheers.....got lots of reading to do.

17. Nov 5, 2007

### viet_jon

got a little backround, but a little stuck.

0 = ( 7.8 x 10² m/s ) x t + ½ x (-9.8 m/s²) x t²

using

should be ?

a= -4.9 m/s² t²
b= 7800 m/s
c= ???

what would c be for this formula? would I just leave it out?

18. Nov 5, 2007

### Feldoh

a = -4.9 m/s^2
b = 780 m/s
c = 0 -- since we don't have a "c term"

19. Nov 5, 2007

### viet_jon

wicked!!!!!!!!

it works!!!!!!!!! problem solved...!!!!!!!!!!!!!!!!!!!!!!

20. Nov 6, 2007

### CompuChip

The advantage of that formula is that it always works for a quadratic equation. All you have to do is bring it to the form $ax^2 + b x + c = 0$ by adding/subtracting terms on both sides of the equation, identify the a, b and c and plug them in.

The disadvantage is, as you might have noticed by now, that some formula's are easier to solve without it. For example, you can solve
$$a t^2 + b t = 0$$
by noticing that you can bring a t outside the brackets (as in the problem you originally posted):
$$(a t + b) t = 0 \Rightarrow t = 0 \text{ or } a t + b = 0 \text{ so } t = - b / a$$
or
$$t^2 - c = 0$$
by just bringing the c to the other side and taking square roots.

But if you don't see such a method, the quadratic formula always works.