# Homework Help: Rearranging numbers changes answers?

1. Jun 11, 2014

1. The problem statement, all variables and given/known data
A car starts from rest and travels with a constant acceleration of $3ms^{-2}$, while a bike which is at a distance of 100m away from the car starts with an initial velocity of $5ms^{-2}$ travels with a constant acceleration of $2ms^{-2}$. The displacement traveled by the bike before being overtaken is $x$. Using equations of motion, find,
(i) the time taken for the car to overtake the bike.
(ii) the distance travelled by the bike (x)
(iii) the distance traveled by the car.

2. Relevant equations
$s= ut + \frac{1}{2} at^{2}$

3. The attempt at a solution
So I did using the theory that the displacement of the car will be equal to displacement of the bike +100m.
So here it goes:
$ut+ \frac{1}{2}at^{2}+100 = ut+ \frac{1}{2}at^{2}$

$\frac{1}{2}.3.t^{2} + 100 = 5t + \frac{1}{2}.2.t^{2}$

$\frac{3}{2}t^{2}+100=5t+t^{2}$

$100=5t-\frac{1}{2}t^{2}$

$\frac{1}{2}t^{2}-5t+100=0$

$t^{2}-10t+200=0$

It is after this the problem started. I got 2 different answers. Here it goes:

$t^{2}-10t+200=0$

--> Here I arranged it as -20t + 10t:

$t^{2} - 20t + 10t + 200 = 0$

$t(t-20)+10(t+20)=0$

$(t+10)(t-20)(t+20) = 0$

$t+10 = 0 / t^{2}-20^{2} = 0$

$t=-10 / t=20 s / t=-20s$

-->Here I arranged it as +10, -20, and got different answers >_>

$t^{2} + 10t - 20t + 200 = 0$

$t(t+10) - 20(t-10) = 0$

$(t-20) (t^{2}-10^{2}) = 0$

$t=20s / t=10 / t=-10$

The answer is supposed to be 20. So how come I also got t=10s in my second arrangement? Is there some mistake I can't identify? Thank you very much in advance c:

Last edited: Jun 11, 2014
2. Jun 11, 2014

### vela

Staff Emeritus
You need to review your algebra. You can't factor the way you did. Note that you started with a quadratic expression, but after factoring, you ended up with a cubic expression. Obviously, you did something you're not allowed to do.

To be specific, $t(t-20) + 10(t+20)$ does not equal $(t+10)(t-20)(t+20)$. You made a similar mistake in the second approach as well.

3. Jun 11, 2014

I figured out what my mistake was ^^;
$s=ut+\frac{1}{2}at^{2}-100 = 5t+t^{2}$
instead of a $+100$ because displacement of the car minus100m should be equal to the displacement of the bike.
Thank you!

4. Jun 11, 2014

### BiGyElLoWhAt

ummm... to reiterate what vela said:

firstly you go from
t(t−20)+10(t+20)=0
to
(t+10)(t−20)(t+20)=0

which is nowhere near correct:
(t+10)(t−20)(t+20)=(t+10)(t^2-400) = t^3 +10t^2 -400t - 4000