Recent content by ballerina_tee

:) Thank you very much! p.s I used x=ut+1/2at^2 to find displacement... x=ut+1/2at^2 =(0)+1/2(3.6) (15^2) = 405m the number is little off because answer should've been 427m, did I use the wrong values?

to find c) I would do v=axt =3.6x15 =54m/s to find b) I would just use one of the equation to find displacement. I think I finally got it.

Oops, I edited it. 27.78t=1.8t^2 is what i got... so, to solve for t... 1.8t^2 - 27.78t =0 roots: t=0 and t=15.43

Since velocity isn't changing... the acceleration is constant? therefore a=0? equation for speeder would be Speeder x=ut+1/2at^2 = (27.78)*t+1/2*a*t^2 = 27.78*t since a=0 .. it would cancel out (1/2*a*t^2) ?

Police x=ut+1/2at^2 = 1/2(3.6)t^2 = 1.8 t^2 Speeder x=ut+1/2at^2 = (27.78)*t+1/2*a*t^2 essentially, ut+1/2at^2=ut+1/2at^2 soo basically i have to get a quadraic equation and the one of the roots will be the time? I'll read over all the post more carefully and try this problem...

Since the distance traveled by both the cop and speeder are the same, to find how long it took for the cop to catch the speeder, do you use this equation v2=v1 + a*t , and solve for t? since acceleration and velocity are given in the problem. I'm so used to solving problems that...

[SOLVED] Kinematic motion problem Homework Statement A police car stopped at a set of lights has a speeder pass it at 100 km/h. If the police car can accelerate at 3.6 m/s^2, a) how long does it take to catch the speeder? b) how far would the police car have to go before it catches the...

Thanks everyone for your contribution! It really helped. Torresmido, no problem, I hope you ace that test. Do these questions over and over until it's embedded in your mind. I'm doing grade 11 physics at the moment so I'm learning the basic concepts. My textbook/teacher gives really easy...

player 1 d=vxt = 3.8x7.5 = 28m ?? if you divide it by 2, it would be 14 but why would you divide by 2?

c) player 2 d=vxt =3.1x7.5 = 23m

now that it's a quadratic formula so i solved for t...(finding roots) it gives me 7.5 :) I understand now! Thanks! finally to solve for b) I would do 0.5 X 7.5 (aXt = v) from a=v/t = 3.75 (3.8)

then, you would set the equations equal to each other and solve for t? I don't get how you got x(t) = .5*.5t^2 or was it a typo? Wouldn't it be x(t) =.5t^2 .5t^2 = 26.9t^2 ? :S

Would I have to set up 2 equations ... 1 being quadratic and the other linear? It's been so long I done math, finding intersections. x(t) = x_i+v_it+\frac{1}{2}at^2 since the initial velocity for player 1 is 0... xi and vi*t would cancel? for player 2, initial velocity/speed would...

I'm still sort of lost. The players are running at different speeds... Would this be a start? finding t for the 2nd player. t=d/v =37/3.1 = 11.9 s (that would be the total time it took the player to run 37 m?) EDIT: ^ I just read your post, I understand what you're trying to say...

[SOLVED] Kinematic motion problem. (acceleration, displacement, velocity) Homework Statement 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. (brackets are...