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.