4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration

Learning objectives

—CHA-3.B—Calculate rates of change in applied contexts• CHA-3.B Calculate rates of change in applied contexts.• CHA-3.B.1 The derivative can be used to solve rectilinear motion problems involving position, speed, velocity, and acceleration.• Enduring understanding CHA-3: Derivatives allow us to solve real-world problems involving rates of change.

For straight-line motion, let $s(t)$ be position at time $t$ . Velocity is the instantaneous rate of change of position, acceleration is the instantaneous rate of change of velocity, and speed is the magnitude of velocity.

v(t)=s'(t),\qquad \text{speed}=|v(t)|,\qquad a(t)=v'(t)=s''(t)

$v(t)>0$ : motion in the positive direction; $v(t)<0$ : motion in the negative direction.\n• Velocity and acceleration with the same sign: speed is increasing.\n• Velocity and acceleration with opposite signs: speed is decreasing.\nIf position is measured in meters and time in seconds, velocity uses $\mathrm{m/s}$ and acceleration uses $\mathrm{m/s^2}$ .

Suppose $s(t)=t^3-3t^2+2t$ meters. Then $v(t)=3t^2-6t+2$ and $a(t)=6t-6$ . At $t=2$ seconds, $s(2)=0\ \mathrm m,\qquad v(2)=2\ \mathrm{m/s},\qquad |v(2)|=2\ \mathrm{m/s},\qquad a(2)=6\ \mathrm{m/s^2}.$ The particle is moving in the positive direction and speeding up because $v(2)$ and $a(2)$ are both positive.

Velocity can be negative because it includes direction; speed cannot be negative. When $v(t)=0$ , the particle is momentarily at rest, and the same-sign/opposite-sign test alone does not classify speeding up at that instant—inspect velocity immediately around the time instead.

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—4.3 Rates of Change in Applied Contexts Other Than MotionRead