Friday, 12 September 2014

POSITION ,VELOCITY PROBLEMS AND SOLUTIONS


Problem1 : Find the derivative of f (x) = 3x 4 -2x 2 +5x -1 and evaluate it at x = 2 .
Using the basic calculus rules established in this section, we find that
f'(x) = 12x 3 -4x - 5x -2 andf'(2) = 96 - 8 - 5/4 = 86 + 3/4
Problem2 : Find the velocity and acceleration functions corresponding to the position function x(t) = 3t 2 - 8t + 458 .
v(t) = x'(t) and a(t) = v'(t) = x''(t) , so using our basic calculus rules again we find that
v(t) = 6t - 8 and a(t) = 6
Notice that the acceleration in this case is constant, and that its value is equal to twice the coefficient of t 2 in x(t) .
Problem 3: What happens when a car which is traveling along at constant velocity screeches to a halt?
The velocity of the car decreases rapidly, corresponding to a large negative acceleration (or deceleration) of the vehicle (courtesy of good brakes). While the car was traveling at constant velocity, on the other hand, the acceleration was zero.