Vertical Motion Scenario Solving for Velocity

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SUMMARY

The discussion centers on solving a vertical motion problem involving two agents, Agent 001 and Mr. LateforClass (LFC), with differing accelerations. LFC accelerates at 10 m/s² with an initial velocity of 0 m/s, while Agent 001 accelerates at 16 m/s² after a 0.75-second reaction time. The key equations used include V² - V₁² = 2(a)(d) and d = (v₁)t + 0.5(a)(t²). The solution involves calculating the final velocity of Agent 001 when he catches LFC and determining the distance traveled, with the intersection point identified at approximately 154 m after 4.7 seconds.

PREREQUISITES
  • Understanding of kinematic equations for uniformly accelerated motion
  • Familiarity with initial velocity and acceleration concepts
  • Ability to graph motion scenarios for visual analysis
  • Knowledge of time intervals in motion problems
NEXT STEPS
  • Study kinematic equations in-depth, focusing on V² - V₁² = 2(a)(d)
  • Learn how to graph motion scenarios to visualize intersections
  • Explore the impact of reaction time on motion calculations
  • Practice solving vertical motion problems with varying initial conditions
USEFUL FOR

Students in physics, educators teaching motion concepts, and anyone interested in solving kinematic problems involving acceleration and reaction times.

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Homework Statement


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.


Homework Equations


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


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 I am doing wrong?

Is there any other way i can get a more accurate answer for either time or distance? I can't use the above equations to solve anything without one or the other..
 

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I can't use the above equations to solve anything without one or the other..
You can, if you start at t=0.75 seconds, for example. In addition, you need an initial position there.
 
I solved it! Thanks for the thought though!
 

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