Question

Suppose that instead of moving at a constant speed, the car in VP 2.9.1 begins to accelerate at 1 m/s2 at the moment it passes the school crossing. How much time elapses before the officer catches up to the car?

What is the position of the car when the officer catches up to it?

PART D AND PART E. THANK YOU

VP 2.9.1 Part A A motorist is traveling at a constant velocity of v = 13.0 m/s and passes a school-crossing corner where the speed limit is 10 m/s (about 22 mi/h). A police officer on a motorcycle stopped at the corner then starts off in pursuit with a constant acceleration of a = 5.00 m/s2. Calculate the positions of the police officer and the car at the moment when they have the same velocity. Express your answers in meters separated by a comma. XP, Ic = 16.9,33.8 m Submit Previous Answers Credit granted by instructor; correct answer displayed

Answer 1

(D) Let the offices catches up the car in t seconds. Hence in time 't both car and officer travel same distance Fue con S=ut that S- 13t+cut , 13tto.st-0 For officel ezut that 2. S=0+{(5) +2 = 2.5+2 - from ① and ② 13t+0.5t² = 2.5t² a 12.5.0.5)+² -13tzo = 2t²_13t = tat-13)=0 = 2t=13 at 12- 6:55 t={61575 1. (E) W xe be position of cor u1 + a2 23 (13x6.5+ y be (0532701 = 105.8m 1xc2 106m /