- #1

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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.

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- Thread starter iamjulius
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- #1

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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.

- #2

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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:

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?

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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