Vertical Motion Scenario Solving for Velocity

1. Feb 13, 2013

cassc94

1. The problem statement, all variables and given/known data
Agent 001 is on another mission... to stop Mr. LateforClass's ploy to threaten mankind with a moon based "Space" Laser, which is powered by rezigrene batteries. 001 tracks LFC for several months and finally finds him on the 2nd floor of an old run down factory. LFC jumps out of a window to escape from 001, and accelerates at 10m/s^2 w/ his rocket pack (initial velocity is 0m/s). 001 is directly 20m below LFC, when he too fires his rocket pack after his 0.75s reaction time, and gives chase (initial velocity 0m/s, acceleration of 16m/s^2). Calculate the final velocity of 001 when he catches LFC and also find the distance 001 travels.

2. Relevant equations
V2^2 - V1^2 = 2(a)(d)
d = (v1)t + 0.5(a)(t^2)
d = (v2)t - 0.5(a)(t^2)

3. The attempt at a solution

I solved the same question except with different accelerations/times/velocities. Earlier i solved it with a table showing their distance over each second, which helped me find the time of their equal distance. This scenario is more difficult particularly because of his 0.75second reaction time. I thought to construct a graph to find the time, but when i plugged it into the equation as seen on my page i get 124...which is not even near what my graph indicates.

My graph shows that they intersect at about 154m of height after about 4.7 seconds.

Can anyone see what im doing wrong?

Is there any other way i can get a more accurate answer for either time or distance? I cant use the above equations to solve anything without one or the other..

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2. Feb 13, 2013

Staff: Mentor

You can, if you start at t=0.75 seconds, for example. In addition, you need an initial position there.

3. Feb 13, 2013

cassc94

I solved it! Thanks for the thought though!