• Aug 28th 2009, 05:59 AM
RaphaelB30
I'm stuck on this problem...what did i do wrong?
d(t)=(velocity) (t)+1/2 (acceleration) (t)^2

cycles 4 meters per second
rate of .4 meters per second squared
vertical distance from top to bottom of hill is 25 meters

How long until goes down the hill?

I have...

(1/5)t^2+4t-25

From the quad formula I got 1/4 and -5/2, but the answer is 5 seconds....why?
• Aug 28th 2009, 06:04 AM
Quote:

Originally Posted by RaphaelB30
I'm stuck on this problem...what did i do wrong?
d(t)=(velocity) (t)+1/2 (acceleration) (t)^2

cycles 4 meters per second
rate of .4 meters per second squared
vertical distance from top to bottom of hill is 25 meters

How long until goes down the hill?

I have...

(1/5)t^2+4t-25

From the quad formula I got 1/4 and -5/2, but the answer is 5 seconds....why?

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

$25=4t+\frac{1}{2}(0.4)t^2$

$
0.2t^2+4t-25=0
$

then $t=\frac{-4\pm\sqrt{4^2-4(0.2)(-25)}}{2(0.2)}$

i got t=5 , -25

so its 5 coz time cant be negative .
• Aug 28th 2009, 06:04 AM
Chris L T521
Quote:

Originally Posted by RaphaelB30
I'm stuck on this problem...what did i do wrong?
d(t)=(velocity) (t)+1/2 (acceleration) (t)^2

cycles 4 meters per second
rate of .4 meters per second squared
vertical distance from top to bottom of hill is 25 meters

How long until goes down the hill?

I have...

(1/5)t^2+4t-25

From the quad formula I got 1/4 and -5/2, but the answer is 5 seconds....why?

Note that $\tfrac{1}{5}t^2+4t-25=0\implies t^2+20t-125=0$

The expression factors into $(t-5)(t+25)=0\implies t=5$ or $t=-25$. We disregard the second answer, since $t>0$.