A speed skater moving across frictionless ice at 9.0 hits a 5.1 -wide patch of rough ice. She slows steadily, then continues on at 6.4 .
What is her acceleration on the rough ice?
a = (V^2-Vo^2)/2d.
a = (40.96-81)/10.2 = -3.93 m/s^2.
"a" is negative because she was slowing down.
Well, you could say she's a "skating drama queen" who just loves to make an entrance by dramatically slowing down on that rough ice patch. But let's not judge her too harshly.
To find her acceleration on the rough ice, we can use the equation:
acceleration = (final velocity - initial velocity) / time
Given that her initial velocity is 9.0 m/s and her final velocity is 6.4 m/s, we need to know the time it takes her to slow down.
But sadly, my crystal ball is in the shop today, so I can't determine the exact time for you. However, if you can provide the time it takes her to slow down, I'd be more than happy to calculate her acceleration on that bumpy ice for you!
To find the acceleration on the rough ice, we can use the equation:
\(a = \frac{{v_f - v_i}}{{t}}\)
Where:
\(a\) is the acceleration
\(v_f\) is the final velocity
\(v_i\) is the initial velocity
\(t\) is the time taken to reach the final velocity
From the given information, we know:
\(v_i = 9.0 \, \text{m/s}\)
\(v_f = 6.4 \, \text{m/s}\)
Now, we need to find the time taken to reach the final velocity. Since the acceleration is steady, we can assume that the acceleration is constant, and we can use the equation of motion:
\(v_f = v_i + at\)
Rearranging the equation, we get:
\(t = \frac{{v_f - v_i}}{{a}}\)
Substituting the given values, we have:
\(t = \frac{{6.4 \, \text{m/s} - 9.0 \, \text{m/s}}}{{a}}\)
Now, we can substitute this value of \(t\) into the equation for acceleration:
\(a = \frac{{v_f - v_i}}{{t}}\)
\(a = \frac{{6.4 \, \text{m/s} - 9.0 \, \text{m/s}}}{{\frac{{6.4 \, \text{m/s} - 9.0 \, \text{m/s}}}}{{a}}}\)
Simplifying this equation, we get:
\(a = \frac{{6.4 \, \text{m/s} - 9.0 \, \text{m/s}}}{t}\)
Now, we can substitute the value of \(t\) obtained earlier:
\(a = \frac{{6.4 \, \text{m/s} - 9.0 \, \text{m/s}}}{\frac{{6.4 \, \text{m/s} - 9.0 \, \text{m/s}}}}{{a}}\)
Simplifying this equation further, we have:
\(a = \frac{{6.4 \, \text{m/s} - 9.0 \, \text{m/s}}}{1}\)
\(a = -2.6 \, \text{m/s}^2\)
Therefore, the acceleration on the rough ice is -2.6 m/s^2. The negative sign indicates deceleration.
To find the acceleration on the rough ice, we can use the equation relating acceleration, initial velocity, final velocity, and distance. The equation is:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
We know that the skater's initial velocity on the rough ice is 9.0 m/s, and her final velocity is 6.4 m/s. We can also assume that the distance over which she slowed down and came to a stop is the width of the rough ice patch, which is 5.1 m.
Substituting these values into the equation, we get:
(6.4)^2 = (9.0)^2 + 2a(5.1)
41.6 = 81 + 10.2a
Rearranging the equation, we have:
10.2a = 41.6 - 81
10.2a = -39.4
Finally, we can solve for acceleration:
a = -39.4 / 10.2
a ≈ -3.86 m/s^2
Therefore, the skater's acceleration on the rough ice is approximately -3.86 m/s^2.