# Constant acceleration: How do I solve time.

• September 18th 2010, 03:32 PM
Constant acceleration: How do I solve time.
HOW DO I CALCULATE TIME?
I need help with this, someone said I have to find time and I have clue on how.ave a problem and I can't solve it. Do I have enough info?

Two knights on horseback start from rest 88.0m apart and ride directly toward each other to do battle. Horse A accelleration has a magnitude of 0.300 m/s^2, while Horse B has a magnitude of 0.200 m/s^2.

Relative to Horse A's starting proint, where do the horses collide?

I am not sure what to do here.

Please let me know ASAP, thanks for looking and helping if you do.
• September 18th 2010, 03:41 PM
skeeter
Quote:

HOW DO I CALCULATE TIME?
I need help with this, someone said I have to find time and I have clue on how.ave a problem and I can't solve it. Do I have enough info?

Two knights on horseback start from rest 88.0m apart and ride directly toward each other to do battle. Horse A accelleration has a magnitude of 0.300 m/s^2, while Horse B has a magnitude of 0.200 m/s^2.

Relative to Horse A's starting proint, where do the horses collide?

I am not sure what to do here.

Please let me know ASAP, thanks for looking and helping if you do.

note the kinematics equation $\Delta x = v_0 t + \frac{1}{2}at^2$

$\frac{1}{2}(0.3)t^2 + \frac{1}{2}(0.2)t^2 = 88$

solve for $t$ ... then determine their individual distances traveled
• September 18th 2010, 03:48 PM
yeKciM
Quote:

HOW DO I CALCULATE TIME?
I need help with this, someone said I have to find time and I have clue on how.ave a problem and I can't solve it. Do I have enough info?

Two knights on horseback start from rest 88.0m apart and ride directly toward each other to do battle. Horse A accelleration has a magnitude of 0.300 m/s^2, while Horse B has a magnitude of 0.200 m/s^2.

Relative to Horse A's starting proint, where do the horses collide?

I am not sure what to do here.

Please let me know ASAP, thanks for looking and helping if you do.

hm... perhaps no need to go into calculating time ... it's obvious that horse B have for 1/3 less acceleration than horse A .... so meaning that horse A will go 2/3 of the way and horse B just 1/3 of the way before collision :D (88/3) * 2 = distance from starting point of the horse A to collision point :D
• September 18th 2010, 04:08 PM
I don't understand the equation, why did you use it twice the end part of the original equ. I did not know you can do that??
I see below that YeKciM using the quadratic equation, not sure how that works. We have not gotten to that in class yet.

Please explain why the dbl use there??
• September 18th 2010, 04:14 PM
skeeter
the sum of their distances traveled when they collide is 88 m
• September 18th 2010, 04:20 PM
yeKciM
Quote:

I don't understand the equation, why did you use it twice the end part of the original equ. I did not know you can do that??
I see below that YeKciM using the quadratic equation, not sure how that works. We have not gotten to that in class yet.

Please explain why the dbl use there??

you probably shouldn't do it like i wrote there, because that's based on just logic... (because skeeter already gave you proper answer) but it should be the same ... about 58.666 m :D

but as for you need this for class go with skeeter's answer :D basically what he just wrote you there is that both horses together are going to travel 88m, so based on there accelerations they will travel different distances (one more than another) ...

Edit : sorry skeeter didn't see you post... for some reason refreshing the page didn't do me any help there...
• September 18th 2010, 04:21 PM