A tailback initially running at a velocity of 6.7 m/s becomes very tired and slows down at a uniform rate of 0.68 m/s2. How fast will he be running after going an additional 6.1 meters?
To find the final velocity of the tailback, we can use the equation of motion:
v^2 = u^2 + 2as
where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- s is the displacement
Given:
Initial velocity, u = 6.7 m/s
Acceleration, a = -0.68 m/s^2 (negative because the tailback is slowing down)
Displacement, s = 6.1 m
We want to find the final velocity, v.
Rearranging the equation, we have:
v^2 = u^2 + 2as
Substituting the given values:
v^2 = (6.7 m/s)^2 + 2 * (-0.68 m/s^2) * 6.1 m
v^2 = 44.89 m^2/s^2 - 8.27 m^2/s^2
v^2 = 36.62 m^2/s^2
Taking the square root of both sides:
v = ā36.62 m^2/s^2
v ā 6.05 m/s
Therefore, the tailback will be running at a speed of approximately 6.05 m/s after going an additional 6.1 meters.