Asked by Luis
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)?
Answers
Answered by
Henry
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.
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.
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