Bus and Sports Car Kinematics Problem

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SUMMARY

The kinematics problem involves a bus traveling at 15.0 m/s east, decelerating at 0.5 m/s² west, and a sports car initially moving at 6.0 m/s east, accelerating at 2.5 m/s² east. To determine when the sports car catches up to the bus, the equations of motion for both vehicles are set equal, leading to the simultaneous equations: d = 15t + 1/2(-0.5)t² for the bus and d = 6t + 1/2(2.5)t² for the car. Solving these equations reveals that the sports car catches the bus after 6 seconds.

PREREQUISITES
  • Understanding of kinematic equations, specifically vf² = vi² + 2ad and d = Vit + 1/2at².
  • Basic knowledge of acceleration and deceleration concepts.
  • Ability to solve simultaneous equations.
  • Familiarity with vector components in motion (east and west directions).
NEXT STEPS
  • Study the derivation and application of kinematic equations in one-dimensional motion.
  • Learn how to solve simultaneous equations in physics problems.
  • Explore the effects of acceleration on different types of vehicles in motion.
  • Investigate real-world applications of kinematics in automotive engineering.
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Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of motion problems involving multiple objects.

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Homework Statement


Question reads: A bus traveling at 15.0m/s [E], started to slow down with an acceleration of 0.5m/s^2[W] as soon as it was abreast of a sports car. At the same instant, the sports car had a velocity of 6m/s [E] and was accelerating at 2.5m/s^2 [E]. At what time does the sports car catch up to the bus?

Homework Equations


vf^2 = vi^2 + 2ad
d = Vit + 1/2at^2


The Attempt at a Solution



d = 15t + 1/2(-0.5)t^2 <-- bus
d = 6t + 1/2(2.5)t^2 <-- car

these are the two equations that i have and will i use these two equations to solve for time ? or distance?

im very confused at this point. ( putting two different object's speed together)

please help me further guide me with this problem.
 
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Hi ya,

The equation you have used is correct - remember, that if the car has caught up with the bus, then d will be the same for both. Hope this helps :)
 


The question sounds so confusing to me :S If the bus and the car are abreast, then the car has already caught up with the bus..

Anyway, usually for this kind of thing, you have 2 equations, use the variable that is in common with both of them, equate them to each other and solve
 


d = 15t + 1/2(-0.5)t^2 <<--- for the bus

d = 6t + 1/2(2.5)t^2 <<-- for the sport car.


Solve the simultaneous equation and you had it.

I hope I'm right @_@.

My answer is 6seconds.
 

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