# ch02-p085 - 85 We denote tr as the reaction time and tb as...

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and a vv vv vx vx =− 1 2 02 01 2 01 02 2 02 1 01 2 . (a) 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 find 22 02 1 01 2 01 02 02 01 (13.4 m/s) (56.7 m) (22.4 m/s) (24.4 m) 0.74 s. ( ) (22.4 m/s)(13.4 m/s)(13.4 m/s 22.4 m/s) r t vv v v == = −− (b) In a similar manner, 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 gives 2 02 01 01 02 02 1 01 2 1 1 (13.4 m/s)(22.4 m/s) (22.4 m/s)(13.4 m/s) 6.2 m/s . 2 2 (13.4 m/s)(56.7 m) (22.4 m/s)(24.4 m) a The magnitude of the deceleration is therefore 6.2 m/s 2 . Although rounded off 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). 85. 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
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