a car catching up to another car


by Whitishcube
Tags: catching
Whitishcube
Whitishcube is offline
#1
Sep25-10, 05:34 PM
P: 83
1. The problem statement, all variables and given/known data
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?


2. Relevant equations
kinematic equations


3. 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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Whitishcube
Whitishcube is offline
#2
Sep25-10, 06:36 PM
P: 83
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?
Tom83B
Tom83B is offline
#3
Sep25-10, 06:48 PM
P: 46
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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