Determine the uniform deceleration of car A

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SUMMARY

The discussion focuses on calculating the uniform deceleration of car A, which is initially traveling at 8.89 m/s while car B is at 6.67 m/s, with a distance of 60 meters separating them. After 45 seconds of braking, the two cars collide, leading to the conclusion that the deceleration of car A is 0.0395 m/s². The calculations involve using the equation of motion s = ut + 1/2at² for both cars and determining their positions over time. The final result confirms that the maximum deceleration for car A, ensuring a collision, is indeed 0.0395 m/s².

PREREQUISITES
  • Understanding of kinematic equations, specifically s = ut + 1/2at²
  • Knowledge of relative motion and collision concepts
  • Familiarity with basic algebra and solving equations
  • Ability to interpret and manipulate units of measurement (m/s, m/s²)
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  • Study advanced kinematics in physics, focusing on motion under uniform acceleration
  • Learn about collision analysis in two-dimensional motion scenarios
  • Explore the implications of deceleration on vehicle safety and stopping distances
  • Investigate real-world applications of motion equations in automotive engineering
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Students studying physics, automotive engineers, and anyone interested in understanding vehicle dynamics and collision analysis.

teng125
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cars A and B are d=60m apart and traveling at u(A)=8.89 and u(B)=6.67.knowing that 45s after driver A apply his brake to avoid overtaking B,the two cars collide,determine the uniform deceleration of car A.

i have tried to form eqns using s=ut + 1/2a(t^2) for both cars and substitute into [car A-car B=60] and can't find the answer.
the answer is 0.0395.

pls help...thanx...
 
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If the units of velocities of the cars is m/s, I am geting 0.1965 m/s/s.MP
 
Take the initial position of B to be x= 0. Its position after t seconds is given by xB= 6.67t m. (assuming that by "u(B)= 6.67" you mean "the speed of B is 6.67 m/s. Please be complete.) A's position when t= 0 is x= -60 m. 45 s later, A's position is xA(45)= 8.89(45)= 400.05 m. If we let "a" be A's decceleration, then A's position at time t, t> 45, is given by
xA(t)= 400.05+ 8.89(t-45)- (a/2)(t-45)2.

Just given the information that "the two cars collide" you can't determine A's decceleration. You can calculate the "greatest decceleration so that they collide"- that is, the largest value for a so that xA(t) and xB(t) are equal for some value of t.
 
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i also got the same answer as urs in threat 2
 

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