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.