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1D Motion

Physics
Summary

Problems for Position, Velocity, and Acceleration in One Dimension 2

Summary Problems for Position, Velocity, and Acceleration in One Dimension 2

Problem : Find the derivative of f (x) = 3x4 -2x2 +5x-1 and evaluate it at x = 2.

Using the basic calculus rules established in this section, we find that

f'(x) = 12x3 -4x - 5x-2 andf'(2) = 96 - 8 - 5/4 = 86 + 3/4

Problem : Find the velocity and acceleration functions corresponding to the position function x(t) = 3t2 - 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 t2 in x(t).

Problem : 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.