Question

A tailback initially running at a velocity of 6.6 m/s becomes very tired and slows down at a uniform rate of 0.55 m/s2. How fast will he be running after going an additional 6.2 meters?

Answer 1

Vf^2=Vi^2+2ad

a= -.55m/s^2, d=6.2, Vi=6.6, solve for Vf

Answer 2

[AI-generated answer — Explain Bot]

To find the final velocity of the tailback after traveling an additional distance of 6.2 meters, we can use the equation for uniformly accelerated motion:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance

Given:
u = 6.6 m/s (initial velocity)
a = -0.55 m/s^2 (acceleration, negative because it is slowing down)
s = 6.2 m (additional distance)

First, let's calculate the displacement (change in distance) using the equation:

s = ut + 0.5at^2

Rearranging the equation:

s = ut + 0.5at^2
6.2 = (6.6)(t) + 0.5(-0.55)(t^2)
6.2 = 6.6t - 0.275t^2

Now, we need to solve this quadratic equation. Let's rearrange it to the standard quadratic form:

0.275t^2 - 6.6t + 6.2 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Where:
a = 0.275
b = -6.6
c = 6.2

t = [ -(-6.6) ± √((-6.6)^2 - 4(0.275)(6.2)) ] / [2(0.275)]

Simplifying the equation:

t = (6.6 ± √(43.56 - 6.82)) / 0.55
t = (6.6 ± √36.74) / 0.55

Now, we can calculate the two possible values of t:

t1 = (6.6 + √36.74) / 0.55
t1 = (6.6 + 6.063) / 0.55
t1 = 12.663 / 0.55
t1 ≈ 23.024 s

t2 = (6.6 - √36.74) / 0.55
t2 = (6.6 - 6.063) / 0.55
t2 = 0.537 / 0.55
t2 ≈ 0.976 s

Since the tailback is slowing down, we can disregard the negative value of time. Therefore, the additional time taken is approximately 0.976 seconds.

Now, substitute this time value back into the original equation to find the final velocity:

v = u + at
v = 6.6 + (-0.55)(0.976)
v ≈ 6.1 m/s

Therefore, the tailback will be running at approximately 6.1 m/s after traveling an additional 6.2 meters.