A double speed trap is set up on a freeway. One police cruiser is hidden behind

Question

A double speed trap is set up on a freeway. One police cruiser is hidden behind a billboard, and another is some distance away under a bridge. As a sedan passes by the first cruiser, its speed is measured to be 100.1 mph. Since the driver has a radar detector, he is alerted to the fact that his speed has been measured, and he tries to slow his car down gradually without stepping on the brakes and alerting the police that he knew he was going too fast. Just taking the foot off the gas leads to a constant deceleration. Exactly 7.00 s later the sedan passes the second police cruiser. Now its speed is measured to be only 69.5 mph, just below the local freeway speed limit.
(a) What is the value of the deceleration? (Take the direction the sedan is going to be the positive direction. Indicate the direction with the sign of your answer.)
(b) How far apart are the two cruisers?

Explanation / Answer

I'm going to convert everything to meters and seconds a = delta v / delta t 100.1 mph = 44.749 m/s 69.5 mph = 31.07 a = (v2 - v1) / t = (31.07 - 44.749) / 7 = -1.954 m/s2

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