## Drag Race Problem

 A Drag Racer accelerates at a uniform rate from it's starting point. It travels the last 1/4 of the distance from the starting point to the finish line in 3 seconds. How long did it take to travel the entire distance from starting point to finish line? This uses the Quadradic Formula: A*X*X + B*X + C = 0 => X = (-B + Sq-Rt(B*B - 4*A*C))/(2*A) => X = (-B - Sq-Rt(B*B - 4*A*C))/(2*A) Enter: Enter .25 = 1/4 of distance here = y1 Enter: Enter 2.5 seconds here = y2 Enter: Enter 2640 = 1/2 mile ** (1320 = 1/4 mile) ** (2640 = 1/2 mile) ** (3960 = 3/4 mile) = y3 Enter: Enter 3200 lbs = weight of the car Enter: Enter .07 = Friction of Car on the Road Answer: b1 = 1 - y1 = .75 Answer: b2 = 1/b1 = 1.333333 Answer: b3 = b2 - 1 = .3333333 Answer: b4 = (2*y2)*(b2)*(-1) = -6.66666 Answer: b5 = y2*y2*b2 = 8.33333 Answer: b6 = b4*b4 - 4*b3*b5 = 33.33333 Answer: b7 = Sq-Rt(b6) = 5.77350269 Answer: b8 = (-b4 + b7)/(2*b3) = Time to travel the whole distance = 18.660254 seconds Answer: b9 = (-b4 - b7)/(2*b3) = 1.33974596 Answer: b10 = y3/b8 = V2 = Ft/Sec = Final Speed = 141.47717 ft/sec Answer: b11 = b10/b8 = a = V2/T = 7.581738 ft/sec/sec Answer: b12 = 88/b11 = Time to Reach 60 mph = Seconds = 11.60683 seconds Answer: b13 = b10*(60/88) = V2 = Final Speed = mph = 96.4617 mph Answer: b14 = y4*y5 = 3200*.07 = 224 lbs to roll car Answer: b15 = y4/32 = 100 slugs = mass of the car Answer: b16 = b15*b11 = ma = Force = 758.173888 lbs Answer: b17 = b14 + b16 = 758.17 + 224 = 982.1738 = Total Force pushing the car Answer: b18 = (88*88)/(2*b11) = (V2*V2)/(2*a) = X = Distance traveled to reach 60 mph = 510.7007 ft Answer: b19 = b17*b18 = Ft-Lbs = Force times Distance = F*X = Work Done = 501596.9758 ft-lbs Answer: b20 = b19/b12 = (Ft-Lbs)/Second = Power = 43215.65108 (Ft-Lbs)/Second Answer: b21 = b20/550 = HP = Horse-Power (550 Ft-Lbs per second = 1 HP) = 78.5739 HP Developed A helium baloon carrying a baseball goes up to a height h feet where it pops and releases the ball. The ball travels the last 3000 feet in 5 seconds. Assume no friction and acceleration due to gravity to be 32 ft/sec/sec. How high was the ball when it dropped? Enter: Enter 3000 feet here Enter: Enter 5 seconds here Answer: V2 - V1 = a*T => 32*5 = 160 ft/sec Answer: V2*V2 - V1*V1 = 2*a*X = 2*32*3000 = 192,000 Answer: V2 + V1 = 192000/160 = 1200 ft/sec Answer: V2 = (1200 + 160)/2 = 680 ft/sec Answer: V1 = V2 - 160 = 680 - 160 = 520 ft/sec Answer: V2 = 8*(SqRt(h)) => h = (V2/8)*(V2/8) = (680/8)*(680/8) = 7225 ft high when dropped ## Horizon Problem  