+25|3546|BEHIND YOU!
Two rugby players are running towards each other. They are 37 m apart. If one is accelerating from rest at 0.5 m/s^2 and the other was already moving at 3.1 m/s and maintains her speed,
a) how long before they crunch together?
b) how fast was the accelerating player going?
c) how far each player ran?

DONT get the damn question....any help would be appreciated.....Motion and Forces sucks big time
I'm moving to Brazil
you ll find ur answer there … 07482.html

Last edited by blademaster (2008-01-11 18:13:19)

The Power of Two
+188|3902|Sydney, Australia
a) OK, let the distance travelled by person A in t seconds be 1/2 * 0.5m/s/s * t^2 (formula). Distance travelled by person B is 3.1*t. Adding them gives you a quadratic equation: t^2 + 12.4*t - 148 = 0. Solving this, you get t = 7.45 seconds.

b) Using the equation v = u + a*t where v is final velocity, u is initial velocity, a is acceleration and t is time, you get: v = 0 + 0.5 * 7.45; v = 3.73m/s

c) Person A: using equation distance = u*t + 1/2*a*t^2 you get distance = 13.88m.
    Person B: distance = speed * time: distance = 23.1m

Last edited by Vub (2008-01-11 23:55:29)

+151|3636|NSW, Australia

theyre chicks!


Lolz... Fizix.
Parcel of ol' Crams

Nappy wrote:

theyre chicks!

lol thats what i was thinking lol

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