Question

A skier starts from rest down a slope 500.0 m long. The skier accelerates at a constant rate of 2.00 m/s2.

What would the velocity of the skier at the bottom of the slope be.

Answer 1

Answer on Question #48589 – Physics – Mechanics | Kinematics | Dynamics

1. A skier starts from rest down a slope $500.0\,\mathrm{m}$ long. The skier accelerates at a constant rate of $2.00\,\mathrm{m/s^2}$. What would the velocity of the skier at the bottom of the slope be.

$$l = 500\,m$$ $$a = 2\,\frac{m}{s^2}$$ $$v = ?$$ $$\text{Solution.}$$

The path of a body during a uniform accelerated motion (with a constant acceleration $a$) can be expressed by the initial and final velocities: $l = \frac{v_{1x}^2 - v_{0x}^2}{2a_x}$, $l = \frac{v^2 - 0^2}{2a}$.

The velocity of the skier at the bottom of the slope will be $\boxed{v = \sqrt{2la}}$.

Let check the dimension: $[v] = \sqrt{m \cdot \frac{m}{s^2} = \frac{m}{s}}$.

Let evaluate the quantity: $v = \sqrt{2 \cdot 500 \cdot 2} = 44.7\left(\frac{m}{s}\right)$.

Answer: $44.7\,\frac{m}{s}$.

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