Suppose the sprinter continues to run at 6m/s until the end pf the race. After crossing the finish line the sprinter decelerates at 2m/s^2.a) how long it takes her to stop b) how far does the sprinter travel while stopping?

vf^2=vi^2+2ad

0=6^2+2*(-2)d
solve for distance d.

how long?
avg speed * time= distance
3 m/s * time=d
time= d/3

Reason for 3 m/s

average speed = (6+0)/2

which is 3

To find the answers to the given questions, we need to use the equations of motion.

Let's start by answering part (a) of the question: How long does it take for the sprinter to stop?

Given:
Initial velocity, u = 6 m/s
Deceleration, a = -2 m/s^2 (negative sign indicates deceleration)
Final velocity, v = 0 m/s (since the sprinter stops)

We can use the formula for acceleration:

v = u + at

Rearranging the formula and solving for time (t):

t = (v - u) / a

Substituting the given values:

t = (0 - 6) / (-2)
t = 6 / 2
t = 3 seconds

Therefore, it takes 3 seconds for the sprinter to stop.

Now let's move on to part (b) of the question: How far does the sprinter travel while stopping?

To find the distance traveled, we can use the formula:

s = ut + (1/2)at^2

Given that the initial velocity u = 6 m/s, the time taken t = 3 seconds, and the deceleration a = -2 m/s^2, we can substitute these values into the formula:

s = (6)(3) + (1/2)(-2)(3)^2
s = 18 - 9
s = 9 meters

Therefore, the sprinter travels a distance of 9 meters while stopping.

In summary:
(a) It takes 3 seconds for the sprinter to stop.
(b) The sprinter travels a distance of 9 meters while stopping.