Students fail to see acceleration as the ratio Dv/Dt.

Below is a diagram of a train that uniformly accelerates from rest to point A, and then travels at a constant speed. Students are asked to calculate the acceleration of the train before reaching point A using the distances and times given. Do not provide them with the kinematic equations.

 

Since the train is travelling at a constant speed after point A, we can find that speed by divided the distance covered by the time interval it took (technically an average), 10 m/s = 40 m / 4 s. Thus we know that at point A the train is travelling at 10 m/s. So the train accelerated from rest to 10 m/s in 10 s. Since acceleration is the ratio of the change in the velocity of the object divided by the time interval it took to undergo that change, the train accelerated at a rate of 1 m/s/s = 10 m/s / 10 s.