You are arguing over a cell phone while trailing an unmarked policecar by 31 m.

Question

You are arguing over a cell phone while trailing an unmarked policecar by 31 m. Both your car and thepolice car are traveling at 100 km/h.Your argument diverts your attention from the police car for1.5 s (long enough for you to look atthe phone and yell, "I won't do that!"). At the beginning of that1.5 s, the police officer begins emergency braking at 5m/s2. (a) What is the separation between the two carswhen your attention finally returns?
1 m

(b) Suppose that you take another 0.4 s to realize your danger andbegin braking. If you too brake at 5 m/s2, what is yourspeed when you hit the police car?
2 km/h (a) What is the separation between the two carswhen your attention finally returns?
1 m

(b) Suppose that you take another 0.4 s to realize your danger andbegin braking. If you too brake at 5 m/s2, what is yourspeed when you hit the police car?
2 km/h

Explanation / Answer

v= 105 -105 =0 a = 5m/s2 t = 1.5sec Distance travelled =at2/2 (As initial relative velocity is zero )          = 5* 1.5*1.5 / 2 =5.625 m   Separation =35 - 5.625 = 29.375 m
b )      Againafter 1.5 +0.4 sec = 1.9 sec     Separation after 1.9 sec = 35 - 5* 1.9*1.9 / 2 = 35 -9.025 = 25.975 m      car's velocity = at = 5* 1.9 = 9.5 m/s or 9.5 *18/ 5 = 34.2 km/h     So time to strike after 1.9 sec = 25.975 / 9.5 = 2.73sec   Now we will think in terms ofabsolute velocity    velocity of police car after (1.9 + 2.73) sec = u-at = 29.16 - 5*4.63 = 6.01 m/s    velocity of car after 2.73 sec of retardation = 29.16 -5*2.73 = 15.51 m/s    so thevelocity of striking will be 15.51 m/s

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