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One dimensional kinematics

  • Thread starter iamjulius
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  • #1
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A car traveling @ 120 km/h passes by a parked police car. If it takes 5 seconds to start the police car, which then accelerates @ 3 m/s^2 to a maximum speed of 150 km/h, how far does the police car travel in overtaking the speeding car, which maintains a speed of 120 km/h?

The 5 seconds is giving me trouble... I assume they will have equal distance (change in X) when the police car overtakes the speeding car, so I set them equal to each other and solved for time...I am stumped.
 

Answers and Replies

  • #2
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Did you remember to add the distance that the bandits car travels?

distance traveled by bandit = (5 seconds + t) * 33.3 m/s

The police cars distance is just a simple kinematic equation:
distance traveled by police = (3 m/s^2 * t)

Then set the distances equal. Is that what you did?
 
  • #3
OlderDan
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dmahmoudi said:
Did you remember to add the distance that the bandits car travels?

distance traveled by bandit = (5 seconds + t) * 33.3 m/s

The police cars distance is just a simple kinematic equation:
distance traveled by police = (3 m/s^2 * t)

Then set the distances equal. Is that what you did?
distance traveled by police = (3 m/s^2 * t)

No it's not, but if you get this distance correct and set the two distances equal you will find the solution to the problem. The distance the police car travels has two contributions: the constant acceleration interval and the constant velocity interval after it reaches top speed. Your expression does not give either of these distances; it gives the velocity of the police car at any time during the constant acceleration interval.
 

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