A screwdriver is drop down an elevator shaft and exactly 5.5 seconds later the sound of the screwdriver hitting the bottom of the shaft is heard. How deep is the shaft? We are also given that the distance the screwdriver falls is given by the equation s = (1/2)gt^2 where t is the time to hit the bottom of the shaft in seconds and g is the acceleration due to gravity = 32 ft/sec-sec. Also the speed of sound = 1,100 ft/sec
Let d = the shaft depth T= total time for the screwdriver to fall and the sound to reach us ts = time for the sound to reach us once the screwdriver hits bottom T = t + ts = 5.5 d= 16t^2 = 1,100 ts let's eliminate ts by noting the ts = 5.5 - t. so we have; 16t^2 = 1,100(5.5-t) or 16t^2 + 1,100t - 6,050 = 0 Using the quadratic formula we find that t = 5.12 sec. Note that there is another solution, but it is negative, which does not make sense for this problem. Now we can find d = 16*5.12^2 = 419.43 ft
Apply the second derivative test to find the local maxima or local minimum of the following function: f(x) = x^2 - 4x + 3
f'(x) = 2x - 4 which = 0 when x = 2 and when x = 2, f(x) = -1 f"(x) = 2 which is positive, so (2,-1) is a minimum
Rick traveled twice as fast as Sean. Thus Rick could travel the 320 miles to the reef in only 2 hours less than it took Sean to travel the 240 miles to Jane's house. Find the rates (speeds) and times for both boys.
Let Rr = Rick's rate; Rs = Sean's rate; tr = Rick's time; ts = Sean's time Given: Rs = 2Rs tr = 320/Rr ts = 240/Rs tr = ts -2 Solving this set of simultaneous equtions yields: Rs = 80 mph; Rs = 40 mph; tr = 4 hours; ts = 6 hours