A car catching up to another car

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A car traveling at 95 km/hr is 110 meters behind another car moving at 75 km/hr, and the goal is to determine how long it takes for the faster car to catch up. The discussion revolves around using kinematic equations, specifically setting the initial and final positions for both cars to solve for the time needed. The approach involves considering the relative velocity and distance between the two cars, with a suggestion to convert speeds to meters per second for accuracy. The final calculated time for the faster car to catch up is approximately 19.8 seconds, confirming the method used is valid. This problem illustrates the application of kinematic equations in relative motion scenarios.
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Homework Statement


a car moving 95 km/hr is 110 meters behind another car going 75 km/hr. how long does it take for the faster car to catch up?


Homework Equations


kinematic equations


The Attempt at a Solution



so i manipulated the kinematic equations to yield an expression for the change in time, but my problem is finding which is initial velocity and which is final. do i have to consider the motion of both cars SEPARATELY? or maybe take the difference of the two velocities and use that as the velocity for the change in distance? any help here would be greatly appreciated. (i also note that the velocities must be converted to m/sec.)
 
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actually i think i might have gotten it. i used the equation

final position = initial position + (init. velocity)(change in time) + (0.5)(accel)(change in time)^2

for both cars, setting acceleration to zero for both.
for the car i set its initial position as zero and final as 110 meters. then for the truck i set BOTH finial and initial position to 110. from there i set the two equations equal to each other and solved for change in time. my final answer is 19.798 seconds. is this the right way to go about this problem?
 
All you have to do is count how long would it take if both cars were by 75km/h slower. So use this in the equation v=s/t...t=s/v.
The result I got was exactly 19.8, so I'm not sure where you got the -0.002. If it's from conversion of speeds to m/s, I think it's easier just to consider the distance as km (110m=0.11km), get the result in hours and multiply by 3600 to get it in seconds.
 
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