We assume the deceleration being constant over time (the motion from the initial speed `v_0 =12 m/s` until the boat reaches the log is uniform decelerated).

The equation that relates the space traveled to the initial speed, acceleration and time is

`s = v_0*t +(a*t^2)/2`

The data in text are `s = 24 m` and `t=4 s` . Therefore

`a = (s-v_0*t)*2/t^2 = (24-12*4)*2/4^2 = -3 m/s^2`

The speed of the boat when it reaches the log is

`v =v_0 +a*t =12 -3*4 =0 m/s`

This means that the boat stops exactly at the contact with the log (just in time).

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