Motion with Constant Acceleration

Kinematics is the study of motion with constant acceleration. We will begin our study of mechanics with translational kinematics, the motion of point particles in one, two, and three dimensions.

Three equations govern kinematics. Each of them can be derived from the definitions of position, displacement, velocity, and acceleration, but they are used so frequently that it is useful to memorize them. They are as follows:

  • v_f=v_i+at
  • x=(1/2)at^2+v_i t
  • ||v_f||^2=||v_i||^2+2ax

When we study motion in multiple dimensions, it is important to remember that each dimension is independent of the others. For example, in the case of a projectile moving in a parabolic path, the up-down (y) direction is the only one affected by the acceleration of gravity a_g=-9.80 m/s^2.

Let’s consider the case of a particle launched upwards with an initial velocity v_i=20.0 m/s. The particle will move in one direction only. We’ll consider the following questions:

  1. How long does the particle take to reach the top of its trajectory?
  2. What is its height at the top of its trajectory?
  3. What will be its velocity as it hits the ground?
  4. How long does the particle take to fall from the top of its trajectory?

In answering these questions, we adopt the convention that the upward direction is positive. Therefore a_g=-9.80 m/s^2 and v_i=20.0 m/s. We want to find t1, the time taken to reach its maximum height; t2, the time taken from the particle’s maximum height until it hits the ground, and v_f, the particle’s velocity when it hits the ground. We know that v=0 when the particle is at the top of its path.

To answer the first question, we use the first equation. We plug in 0 m/s=20.0 m/s + (-9.80 m/s^2)t1 and solve for t1, finding t1=2.04 s.

To answer the second question, we use the second equation. We plug in y=(1/2)(-9.80 m/s^2)(2.04 s)^2+(20.0 m/s)(2.04 s), finding y=20.4 m.

To answer the third question, we use the third equation. We plug in ||v_f||^2=(0. m/s)^2+2(-9.80 m/s^2)(20.4 m), finding v_f=-20.0 m/s. (We could have equivalently derived this from the conservation of energy; note that the final speed is the same as the initial speed, since the particle begins and ends at the same height.)

To answer the final question, we return to the first equation, plugging in -20.0 m/s = 0. m/s + (-9.80 m/s^2)t2. Solving for t2, we find t2=2.04 s.

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