Integrating Acceleration to Find Motion and Stopping Distance

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How are acceleration and velocity related?

Instantaneous acceleration is the time derivative of velocity: $a(t)=\frac{dv(t)}{dt}$. Integrating acceleration gives velocity, up to a constant.

How do you find velocity from a known acceleration function?

Calculate $v(t)=\int a(t)\,dt+C_1$, then use an initial velocity condition to determine $C_1$.

How are velocity and position related?

Instantaneous velocity is the time derivative of position, $v(t)=\frac{dx(t)}{dt}$. Integrating velocity gives position plus a constant.

What is the constant-acceleration velocity equation?

$v(t)=v_0+at$, where $v_0$ is initial velocity, $a$ is constant acceleration, and $t$ is elapsed time.

What is the constant-acceleration position equation?

$x(t)=x_0+v_0t+\frac{1}{2}at^2$, where $x_0$ and $v_0$ are the initial position and velocity.

Quiz questions

1. A motorboat has acceleration $a(t)=-\frac{1}{4}t$ and initial velocity $v(0)=5.0\,\text{m/s}$. With time in seconds, which expression gives its velocity in meters per second?Answer: $v(t)=5.0-\frac{1}{8}t^2$
2. A motorboat's velocity is $v(t)=5.0-\frac{1}{8}t^2$ in meters per second, with time in seconds. When does it first stop?Answer: At approximately $6.3\,\text{s}$
3. A motorboat's position is $x(t)=5.0t-\frac{1}{24}t^3$ meters, with time in seconds. What displacement has it reached when $t=\sqrt{40}\,\text{s}$?Answer: Approximately $21.1\,\text{m}$
4. What does the slope of a position-versus-time graph represent?Answer: The object's velocity
5. In the model $v(t)=5.0-\frac{1}{8}t^2$, what does negative velocity after the first zero indicate?Answer: The model predicts motion in the negative direction.