A boy sledding down a hill accelerates at 1.70 m/s2. If he started from rest, in what distance would he reach a speed of 7.40 m/s?

To find the distance the boy covers while sledding, we can use the kinematic equation:

\(v^2 = u^2 + 2as\),

where:
- \(v\) is the final velocity (7.40 m/s),
- \(u\) is the initial velocity (0 m/s, as the boy started from rest),
- \(a\) is the acceleration (1.70 m/s^2),
- \(s\) is the distance covered.

Substituting the given values into the equation, we have:

\(7.40^2 = 0^2 + 2 \times 1.70 \times s\).

Simplifying the equation, we get:

\(54.76 = 3.4s\).

Now, we need to isolate \(s\) by dividing both sides of the equation by 3.4:

\(s = \frac{54.76}{3.4}\).

Evaluating \(s\), we find:

\(s \approx 16.1\) meters.

Therefore, the boy would reach a speed of 7.40 m/s after approximately 16.1 meters of sledding down the hill.