Determine the uniform deceleration of car A

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To determine the uniform deceleration of car A, the initial positions and velocities of both cars must be established. Car A starts 60 meters behind car B, moving at 8.89 m/s, while car B moves at 6.67 m/s. After 45 seconds of braking, the positions of both cars can be expressed using kinematic equations. The calculations show that the maximum deceleration for car A, ensuring a collision occurs, is approximately 0.0395 m/s². This value is derived from setting the equations for both cars equal at the point of collision.
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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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