A dragster accelerates from rest for a distance of 450m at 14 m/s^2. A parachute is then used to slow it down to a stop if the parachute gives the dragster an acceleration of -7m/s^2 how far has the dragster traveled before stopping?

Answer 1

The distance travelled while accelerating at a rate a=14 m/s^2 is X = 450 m and the maximum velocity attained at that time is V = sqrt(2aX) = 112.2 m/s. The time required to decelerate to a stop at a rate a' = 7 m/s^2 is t = V/a' = 16.0 s

The average velocity while decelerating is
v = V/2 = 56.1 m/s and the distance travelled while declerating is v*t
X' = 56.1 m/s*16 s = 900 m

The total distance travelled when it has stopped is 450 + 900 = 1350 m

Here is an easier way: Since the rate of deceleration is half the rate of acceleration, it takes twice as long to decelerate and one travels twice as far (900 m)while decereating.