Q: Officer Chuck, a former track star at Clemson University, is in pursuit of a suspect that is 500 meters ahead of him. The suspect runs with a speed of 5 meters per second. Conversely, Chuck is running at a speed of 30 meters per second. Assuming that both participants in this sprint run in a straight line, how long will it take Chuck to catch the suspect?
A: The following diagram will be of great assistance in solving this type of problem:
The key to solving this problem lies in the fact that Chuck and Suspect X will meet at some common point after traveling some period of time ( t ). During this period of time, these gentlemen will have moved the following distance:
When Suspect X is captured, he will be located a distance of ( d1 + 500 m ) from Chuck’s starting point. This distance is the distance Chuck will have also covered while running at a higher speed:
( 30 m/s )( t ) = ( 5 m/s )( t ) + 500 m
We are now ready to determine how many seconds it will take for Chuck to tackle the suspect:
( 30 m/s )( t ) – ( 5 m/s )( t ) = 500 m
( 30 m/s – 5 m/s )( t ) = 500 m
( 25 m/s )( t ) = 500 m
t = [ 500 m / ( 25 m/s ) ]
t = 20 s
Bonus: How far will Chuck have run to catch the suspect?
( 30 m/s )( 20 s ) = 600 m
RENAISSANCE PHYSICS 