Posted by Tri on Monday, September 10, 2007 at 3:41am.
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.
Related Questions
Physics - A dragster accelerates from rest for a distance of 450m at 14 m/s^2. A...
Physics - A dragster accelerates from rest for a distance of 450m at 14 m/s^2. A...
physics - Stickman is driving a dragster. He does the quarter mile in 4 seconds ...
physics - Stickman is driving a dragster. He does the quarter mile in 4 seconds...
physics - A(n) 840-kg dragster accelerates on a race track from rest to 85 km/h ...
Physics - A dragster starts from rest and accelerates down the track. Each tire ...
physics - A tortoise and hare start from rest and have a race. As the race ...
Physics - A tortoise and hare start from rest and have a race. As the race ...
physics - An 879-kg (1943 lb) dragster, starting from rest, attains a speed of ...
Physics - A dragster travels 1/4 mile in 6.1 seconds. Assuming the acceleration...
For Further Reading
