A skate boarder traveling at a constant speed of 22 m/s on a road comes to a straight incline with a constant slope (do not use g for incline). While going up the incline, the skater has constant deceleration. If she is 16m up the incline after 12.0 s
A) what is the acceleration of the skater as she skates the incline
B) What is the velocity of the skater after 12s?
C) when does the skater reach her highest point on the slope?
To solve this problem, we can use the kinematic equation for motion along an inclined plane:
\(y = y_0 + v_0t + \frac{1}{2}at^2\),
where:
y is the displacement along the incline (16 m),
y_0 is the initial displacement (0 m),
v_0 is the initial velocity (22 m/s),
t is the time (12.0 s), and
a is the acceleration we need to find.
A) To find the acceleration of the skater as she skates up the incline, we can rearrange the equation:
\(2(y - y_0) = v_0t + at^2\).
Substituting the given values:
\(2(16 - 0) = 22(12.0) + a(12.0)^2\),
\(32 = 264 + 144a\),
Simplifying further:
\(32 - 264 = 144a\),
\(-232 = 144a\),
Dividing both sides by 144:
\(\frac{-232}{144} = a\),
\(a \approx -1.61 \, \text{m/s}^2\).
Therefore, the acceleration of the skater as she skates up the incline is approximately -1.61 m/s^2.
B) To find the velocity of the skater after 12 s, we can use the equation:
\(v = v_0 + at\).
Substituting the given values:
\(v = 22 + (-1.61)(12.0)\),
\(v = 22 - 19.32\),
\(v \approx 2.68 \, \text{m/s}\).
Therefore, the velocity of the skater after 12 s is approximately 2.68 m/s.
C) To find when the skater reaches her highest point on the slope, we can calculate the time when the velocity becomes 0.
Using the equation:
\(v = v_0 + at\),
and substituting the known values:
\(0 = 22 - 1.61t\),
Solving for t:
\(1.61t = 22\),
\(t = \frac{22}{1.61}\),
\(t \approx 13.67 \, \text{s}\).
Therefore, the skater reaches her highest point on the slope after approximately 13.67 s.
To solve this problem, we'll need to use the equations of motion and the principles of kinematics. Let's break down the problem step-by-step.
Step 1: Identify the given information:
- Initial speed (vi) of the skateboarder = 22 m/s
- Displacement (Δx) of the skateboarder on the incline = 16 m
- Time (t) taken to cover the displacement = 12.0 s
Step 2: Solve for the acceleration (a):
We can use the formula of motion: Δx = vi * t + (1/2) * a * t^2. Since we know Δx, vi, and t, we can rearrange the equation to solve for a.
Δx = vi * t + (1/2) * a * t^2
16 = 22 * 12 + (1/2) * a * (12^2)
Simplifying, we get:
16 = 264 + 6a * 144
16 - 264 = 864a
-248 = 864a
a = -248 / 864
a ≈ -0.287 m/s^2
Therefore, the acceleration of the skateboarder as she skates up the incline is approximately -0.287 m/s^2 (negative sign indicates deceleration).
Step 3: Determine the velocity (vf) of the skateboarder after 12 seconds:
To find the velocity, we can use the formula: vf = vi + a * t.
vf = 22 + (-0.287) * 12
vf = 22 - 3.444
vf ≈ 18.556 m/s
Therefore, the velocity of the skateboarder after 12 seconds is approximately 18.556 m/s.
Step 4: Find the time when the skateboarder reaches the highest point on the slope:
At the highest point on the slope, the skater's velocity will be zero. Since we have the initial velocity, final velocity, and acceleration, we can use the formula: vf = vi + a * t rearranged to solve for t.
0 = 22 + (-0.287) * t
-0.287t = -22
t = -22 / (-0.287)
t ≈ 76.63 s
Therefore, the skateboarder reaches her highest point on the slope after approximately 76.63 seconds.
In summary:
A) The acceleration of the skater as she skates up the incline is approximately -0.287 m/s^2 (deceleration).
B) The velocity of the skater after 12 seconds is approximately 18.556 m/s.
C) The skater reaches her highest point on the slope after approximately 76.63 seconds.