Motion problem for constant acceleration

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SUMMARY

The discussion focuses on a physics problem involving a pedestrian running at a maximum speed of 6.0 m/s attempting to catch a bus that accelerates from rest at 1.0 m/s². The pedestrian starts 16 meters behind the bus when the traffic light changes. The key equations for displacement are provided, specifically d = vi * t + 0.5 * a * t² and the more general form d = di + vit + ½at². The main question is whether the pedestrian can catch the bus and, if not, what the closest distance of approach is.

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chroncile
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Homework Statement


A pedestrian is running at his maximum speed of 6.0 m/s trying to catch a bus that is stopped at a traffic light. When he is at 16 m from the bus, the light changes and the bus pulls away from the pedestrian with an acceleration of 1.0 m/s2.

Does the pedestrian catch the bus and, if so, how far does he have to run? (If not, what is the pedestrian's distance of the closest approach?)


Homework Equations


d = vi * t + 0.5 * a * t2


The Attempt at a Solution


I don't understand how to do it :frown:
 
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chroncile said:

Homework Equations


d = vi * t + 0.5 * a * t2

The Attempt at a Solution


I don't understand how to do it :frown:
Show us your best try. :biggrin: What is your equation for the displacement of the bus, from the traffic light, as a function of time t? What is your equation for the pedestrian's displacement from the traffic light as a function of time t? Is there ever a real-valued, positive time t where the displacements are equal? :wink:

Hint: there's a more general equation that you might consider using, instead of the one you quoted (and don't forget the bus and the pedestrian each have their own individual velocities and accelerations):

d = di + vit + ½at2
 

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