Respuesta :
acceleration of the skier on the inclined plane is given by
[tex]a = g sin\theta[/tex]
[tex]a = 9.81 * sin10[/tex]
[tex]a = 1.70 m/s^2[/tex]
now by equation of kinematics
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]15^2 - 3^2 = 2 * 1.7 * d[/tex]
[tex]d = 63.53 m[/tex]
so the length of inclined plane will be 63.53 m
Part b)
Time to reach the bottom will be given by
[tex]v_f - v_i = at[/tex]
[tex]15 - 3 = 1.7 * t[/tex]
[tex]t = 7.06 s[/tex]
so it will take 7.06 s to reach the bottom
The length of the inclined plane is 63.53 m and the time taken by the skier to cover the distance is 7.0588 sec.
Given to us
Velocity, u = 3.0 m/s
Angle of Inclination = 10"
Velocity of skier at the bottom, v = 15 m/s
What is the acceleration of the skier on the inclined plane?
We know that acceleration of the skier on the inclined plane can be given as,
a = g sin(θ)
a = g sin(10)
a= 9.81 x 0.1736
a = 1.70 m/sec²
What is the length of the incline?
We know that according to the third equation of motion
[tex]v^2-u^2=2as[/tex]
Substitute the values,
[tex]15^2-3^2=2\times 1.70 \times s\\225-9=3.4 s\\s = 63.53\rm\ m[/tex]
How long does it take him to reach the bottom?
Using the first equation of motion,
[tex]v-u=at\\\\15-3=1.70t\\t=\dfrac{15-3}{1.70}\\\\t = 7.0588\rm\ sec[/tex]
Hence, the length of the inclined plane is 63.53 m and the time taken by the skier to cover the distance is 7.0588 sec.
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