A boy sledding down a hill accelerates at 2.00 m/s2. If he started from rest, in what distance would he reach a speed of 5.20 m/s?

To find the distance the boy would reach a speed of 5.20 m/s, we can use the kinematic equation:

v^2 = u^2 + 2as

where:
- v is the final velocity (5.20 m/s),
- u is the initial velocity (0 m/s, as he started from rest),
- a is the acceleration (2.00 m/s^2),
- s is the distance traveled.

Rearranging the equation, we have:

s = (v^2 - u^2) / (2a)

Substituting the given values:

s = (5.20^2 - 0) / (2 × 2.00)

Simplifying:

s = (27.04 - 0) / 4.00

s = 27.04 / 4.00

s = 6.76 meters

Therefore, the boy would reach a speed of 5.20 m/s in a distance of 6.76 meters.

Use v²-u²=2aS

where
v=final velocity
u=initial velocity
a=acceleration, and
S=distance travelled.

Solve for S.