p02_037

Fundamentals of Physics,Vol 1 (Chapters 1 - 20)

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37. We denote t r as the reaction time and t b as the braking time. The motion during t r is of the constant- velocity (call it v 0 ) type. Then the position of the car is given by x = v 0 t r + v 0 t b + 1 2 at 2 b where v 0 is the initial velocity and a is the acceleration (which we expect to be negative-valued since we are taking the velocity in the positive direction and we know the car is decelerating). After the brakes are applied the velocity of the car is given by v = v 0 + at b . Using this equation, with v = 0, we eliminate t b from the Frst equation and obtain x = v 0 t r v 2 0 a + 1 2 v 2 0 a = v 0 t r 1 2 v 2 0 a . We write this equation for each of the initial velocities: x 1 = v 01 t r 1 2 v 2 01 a and x 2 = v 02 t r 1 2 v 2 02 a . Solving these equations simultaneously for t r and a we get t r = v 2 02 x 1 v 2 01 x 2 v 01 v 02 ( v 02 v 01 ) and a = 1 2 v 02 v 2 01 v 01 v 2
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Unformatted text preview: 02 v 02 x 1 − v 01 x 2 . Substituting x 1 = 56 . 7 m, v 01 = 80 . 5 km / h = 22 . 4 m / s, x 2 = 24 . 4 m and v 02 = 48 . 3 km / h = 13 . 4 m/s, we Fnd t r = 13 . 4 2 (56 . 7) − 22 . 4 2 (24 . 4) (22 . 4)(13 . 4)(13 . 4 − 22 . 4) = 0 . 74 s and a = − 1 2 (13 . 4)22 . 4 2 − (22 . 4)13 . 4 2 (13 . 4)(56 . 7) − (22 . 4)(24 . 4) = − 6 . 2 m / s 2 . The magnitude of the deceleration is therefore 6 . 2 m/s 2 . Although rounded o± values are displayed in the above substitutions, what we have input into our calculators are the “exact” values (such as v 02 = 161 12 m/s)....
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