A high−speed test vehicle is brought to rest by throwing out a drag chute behind it, causing a constant acceleration of −12.0 m/s2. The vehicle has a velocity of 50.0 m/s when the chute ejects. What is the shortest distance needed for the test vehicle to stop after the chute is ejected?
+37084 When the vehicle stops, it's velocity will be 0 vf = 0
Use vf = vo + a t to calculate 't' vo = original velocity 50 m/s a = -12 m/s^2
then use the 't' in
xf = xo + vot + 1/2 a t^2 to find xf= distance to stop xo = 0 in this instance
