Frank is traveling the posted speed limit of 36 mi/hr in a school zone. A deer darts out between two cars. He instantly hits the brakes and the car decelerates at a rate of 5.0 m/s^2. a) how much stopping distance will he require in (meters)? b) If he was going twice the original speed what would be his stopping distance (meters)?

Question

Henry · Answer

a. Vo=36mi/h * 1600m/mi * 1h/3600s=16 m/s. V^2 = Vo^2 + 2a*d d = (V^2-Vo^2)/2a d = (0-256)/-10 = 25.6 m. b. d = 0-1024)/-10 = 102.4 m.