3.6 Finding velocity and displacement from acceleration  (Page 34/6)

This section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Average and Instantaneous Acceleration we introduced the kinematic functions of velocity and acceleration using the derivative. By taking the derivative of the position function we found the velocity function, and likewise by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, we can work backward and calculate the velocity function from the acceleration function, and the position function from the velocity function.

Kinematic equations from integral calculus

Let’s begin with a particle with an acceleration a (t) is a known function of time. Since the time derivative of the velocity function is acceleration,

we can take the indefinite integral of both sides, finding

where C ₁ is a constant of integration. Since ${\displaystyle \int \frac{d}{dt}v(t)dt}=v(t)$ , the velocity is given by

Similarly, the time derivative of the position function is the velocity function,

Thus, we can use the same mathematical manipulations we just used and find

where C ₂ is a second constant of integration.

We can derive the kinematic equations for a constant acceleration using these integrals. With a ( t ) = a a constant, and doing the integration in [link] , we find

If the initial velocity is v (0) = v ₀ , then

Then, C ₁ = v ₀ and

which is [link] . Substituting this expression into [link] gives

Doing the integration, we find

If x (0) = x ₀ , we have

so, C ₂ = x ₀ . Substituting back into the equation for x ( t ), we finally have

which is [link] .

Motion of a motorboat

Strategy

Solution