Question

Лыжник скатился с горы длиной s1=60 м за t1=15 сек, а затем проехал по горизонтальному участку еще s2=30 м до остановки. На обоих участках движение было равноускоренным. Найдите скорость v1 лыжника в конце спуска и ускорение a2 на горизонтальном участке.

Answer 1

Едем с горы:
S₁=a*t² / 2
a₁=2*S₁/t² = 2*60/15² = 0,53 м/с²
V₁=a₁*t = 0,53*15 ≈ 8 м/с

Катимся по равнине
S₂=V₁²/(2*a₂)
a₂=V₁²/(2*S₂) = 8² / (2*30) ≈ 1,1 м/с²

Answer 2

Calculation of the Skier's Speed at the End of the Descent (v1)

To find the speed of the skier at the end of the descent (v1), we can use the formula for uniformly accelerated motion:

v = u + at

Where: - v is the final velocity - u is the initial velocity - a is the acceleration - t is the time

In this case, the skier starts from rest (u = 0) and travels a distance of s1 = 60 m in a time of t1 = 15 seconds. We need to find the acceleration (a) and the final velocity (v1).

To find the acceleration (a), we can use the formula:

s = ut + (1/2)at^2

Where: - s is the distance traveled - u is the initial velocity - t is the time - a is the acceleration

Rearranging the formula, we get:

a = (2s) / t^2

Substituting the given values, we have:

a = (2 * 60) / (15^2)

Calculating this, we find:

a = 0.533 m/s^2

Now, we can use the formula for final velocity to find v1:

v1 = u + at

Since the skier starts from rest (u = 0), the formula simplifies to:

v1 = at

Substituting the values of a and t1, we have:

v1 = 0.533 * 15

Calculating this, we find:

v1 = 7.995 m/s

Therefore, the skier's speed at the end of the descent (v1) is approximately 7.995 m/s.

Calculation of the Acceleration on the Horizontal Section (a2)

To find the acceleration on the horizontal section (a2), we can use the formula for uniformly accelerated motion:

v^2 = u^2 + 2as

Where: - v is the final velocity - u is the initial velocity - a is the acceleration - s is the distance traveled

In this case, the skier starts from rest (u = 0) and travels a distance of s2 = 30 m. We need to find the acceleration (a2).

Rearranging the formula, we get:

a2 = (v^2 - u^2) / (2s)

Since the skier starts from rest (u = 0), the formula simplifies to:

a2 = v^2 / (2s)

Substituting the value of v1 (which we calculated earlier) and s2, we have:

a2 = (7.995^2) / (2 * 30)

Calculating this, we find:

a2 = 1.598 m/s^2

Therefore, the acceleration on the horizontal section (a2) is approximately 1.598 m/s^2.

Please let me know if there's anything else I can help you with!