CDK3Y
Member
+25|3304|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
blademaster
I'm moving to Brazil
+2,075|3810
you ll find ur answer there
http://www.physicsforums.com/archive/in … 07482.html

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

Vub
The Power of Two
+188|3659|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)

Nappy
Apprentice
+151|3394|NSW, Australia

theyre chicks!

sweeet
Entertayner
Member
+826|3735

Lolz... Fizix.
Tehremos
Parcel of ol' Crams
+128|3573|Somersetshire

Nappy wrote:

theyre chicks!

sweeet
lol thats what i was thinking lol

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