Answer
A minimum acceleration of $15.1~m/s^2$ is required to keep the stuntman on the truck.
Work Step by Step
Let $F_N$ be the normal force from the truck exerted on the stuntman. The normal force from the truck on the stuntman accelerates the stuntman. We can find an expression for the normal force $F_N$:
$F_N = ma$
In order to stay on the front of the truck, the force of friction exerted upward on the stuntman must be equal in magnitude to the stuntman's weight.
$F_N~\mu = mg$
$F_N = \frac{mg}{\mu}$
$ma = \frac{mg}{\mu}$
$a = \frac{g}{\mu} = \frac{9.80~m/s^2}{0.65} = 15.1~m/s^2$
A minimum acceleration of $15.1~m/s^2$ is required to keep the stuntman on the truck.
