Math Exploration 10.1

We have not seen yet how to deal with accelerated motion. It is not too different from unaccelerated motion, but we have to take into account the fact that the speed is not constant. For constant acceleration the distance d traveled by some object in a given time t is:

 

\begin{equation} d=v_0t+\frac{1}{2}at^2 \end{equation}

 

where v₀ is the initial speed of the object and a is its acceleration. It looks almost like the expression we used before, but there is an extra term that takes into account our changing speed (see Going Further 10.1: Accelerated Motion).

 

We can use this expression for an accelerating boat to calculate how far the boat travels while the seagull is flying over it:

• Given: the acceleration of the boat a = 1 m/s²; the time it takes the seagull to fly over is t = 10 s; the boat starts from rest so v₀ = 0
• Find: the distance the boat travels while the seagull is flying over it
• Concept: d = v₀t + (½)at²
• Solution: d = 0 + (½)(1 m/s²)(10 s)² = 50 m.
• So the boat has moved the same distance as in the prior example.