A stuntman drives a dirt bike on a curved track with a radius of 16.9 m. If he starts from rest and accelerates at 1.06 m/s2, at what time will the magnitude of the total acceleration of the bike be 5.74 m/s2? (Hint: the tangential acceleration will stay constant at 1.06 m/s2.)

The total acceleration of an object moving in a circle is given by the vector sum of the radial acceleration (ar) and the tangential acceleration (at). The radial acceleration is given by the equation ar = v^2/r, where v is the velocity and r is the radius of the circle. The tangential acceleration is given by the equation at = dv/dt, where dv is the change in velocity and dt is the change in time.

Given that the total acceleration (a) is 5.74 m/s^2, the tangential acceleration (at) is 1.06 m/s^2, and the radius (r) is 16.9 m, we can use the Pythagorean theorem to solve for the radial acceleration (ar):

a^2 = ar^2 + at^2 => ar = sqrt(a^2 - at^2) => ar = sqrt((5.74 m/s^2)^2 - (1.06 m/s^2)^2) => ar = 5.63 m/s^2

Now we can use the equation for radial acceleration to solve for the velocity (v):

ar = v^2/r => v = sqrt(ar * r) => v = sqrt((5.63 m/s^2) * (16.9 m)) => v = 31.1 m/s

Finally, we can use the equation for tangential acceleration to solve for the time (t):

at = dv/dt => t = dv/at => t = (31.1 m/s) / (1.06 m/s^2) => t = 29.3 s

So, the magnitude of the total acceleration of the bike will be 5.74 m/s^2 after approximately 29.3 seconds.