Physics
31.10.2022 19:49
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Askier is gliding along at 3.0 m/s on horizontal, frictionless snow. he suddenly

Askier is gliding along at 3.0 m/s on horizontal, frictionless snow. he suddenly starts down a 10" incline. his speed at the bottom is 15 m/s. a. what is the length of the incline? b. how long does it take him to reach the bottom?
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dward5823
dward5823
5,0(77 marks)

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

v^2-u^2=2as

Substitute the values,

15^2-3^2=2\times 1.70 \times s\\225-9=3.4 s\\s = 63.53\rm\ m

How long does it take him to reach the bottom?

Using the first equation of motion,

v-u=at\\\\15-3=1.70t\\t=\dfrac{15-3}{1.70}\\\\t = 7.0588\rm\ sec

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.

Learn more about Motion:

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herringalyssa
herringalyssa
4,4(47 marks)

acceleration of the skier on the inclined plane is given by

a = g sin\theta

a = 9.81 * sin10

a = 1.70 m/s^2

now by equation of kinematics

v_f^2 - v_i^2 = 2 a d

15^2 - 3^2 = 2 * 1.7 * d

d = 63.53 m

so the length of inclined plane will be 63.53 m

Part b)

Time to reach the bottom will be given by

v_f - v_i = at

15 - 3 = 1.7 * t

t = 7.06 s

so it will take 7.06 s to reach the bottom

logan12345677885675
logan12345677885675
5,0(59 marks)
When It begins to drop because that when gravity will have its strongest pull on the object.

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