Answer:
Approximately [tex]4.9\; {\rm m\cdot s^{-1}}[/tex] backward, assuming that the canoe was initially stationary.
Explanation:
When an object of mass [tex]m[/tex] is moving at a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of that object would be [tex]p = m\, v[/tex].
With a mass of [tex]m = 67\; {\rm kg}[/tex] and a velocity of [tex]v = 3.3\; {\rm m\cdot s^{-1}}[/tex] forward, the momentum of Fred would be [tex]p = m\, v = 67\; {\rm kg} \times 3.3\; {\rm m\cdot s^{-1}} = 22.11\; {\rm kg \cdot m \cdot s^{-1}}[/tex] (forward.)
Before the jump, the velocity of the boat and Fred were both [tex]0\; {\rm m \cdot s^{-1}}[/tex]. Hence, the momentum of both Fred and the boat would initially be [tex]0\; {\rm kg \cdot m \cdot s^{-1}}[/tex].
Momentum is preserved immediately after the jump. Thus, the momentum of Fred plus the momentum of the boat would still be [tex]0\; {\rm kg \cdot m \cdot s^{-1}}[/tex] after the jump.
Since the momentum of Fred was [tex]22.11\; {\rm kg \cdot m \cdot s^{-1}}[/tex], the momentum of the boat would be [tex](-22.11\; {\rm kg \cdot m \cdot s^{-1}})[/tex] (backward, as the negative sign indicates.)
Rearrange [tex]p = m\, v[/tex] to find velocity in terms of mass [tex]m[/tex] and momentum [tex]p[/tex]:
[tex]\begin{aligned} v &= \frac{p}{m} \\ &= \frac{-22.11\; {\rm kg \cdot m \cdot s^{-1}}}{67\; {\rm kg}} \\ &\approx (-4.9)\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].
In other words, the velocity of the boat would be approximate [tex]4.9\; {\rm m\cdot s^{-1}}[/tex] backwards.
