Question

a skier is moving 8.33 m/s when he starts to slide up a 8.44 frictionless slope. how much time does it take him to come to a stop

Answer 1

The skier will come to a stop in approximately 1.01 seconds.

Step-by-step explanation:

The motion of the skier on the slope can be analyzed using the kinematic equation:
`\(v_f = v_i + at\)`, where
`\(v_f\)` is the final velocity (0 m/s, as the skier comes to a stop),
`\(v_i\)` is the initial velocity (8.33 m/s), a is the acceleration, and t is time. In this case, the acceleration is due to gravity acting along the slope. On an inclined plane, the acceleration can be expressed as

a = g sin θ, where g is the acceleration due to gravity (approximately 9.8 m/s²) and θ is the angle of the slope. In this scenario, the slope is frictionless, so there is no opposing force.

Now, substitute the values into the kinematic equation:
`\(0 = 8.33 + (9.8 \sin θ)t\)`. Solve for t, and you find that t ≈ 1.01 seconds. This means it takes approximately 1.01 seconds for the skier to come to a stop while moving up the 8.44-meter frictionless slope. The positive sign in the equation indicates that the acceleration is acting opposite to the initial velocity, as the skier is moving against the direction of gravity. This aligns with our physical intuition, as we expect the skier to slow down and eventually come to a stop when moving uphill on a slope.

Full Question:

How much time does it take for a skier, moving at 8.33 m/s, to come to a stop while sliding up an 8.44-meter frictionless slope?

Answer 2

It takes approximately
`\(5.776 \, \text{s}\)` for the skier to come to a stop while sliding up the slope.

The skier is moving up a frictionless slope, meaning only the component of gravity parallel to the slope affects the skier's motion. The skier's initial velocity and the slope angle are given.

The acceleration due to gravity can be split into two components: one perpendicular to the slope (which doesn't affect the skier's motion) and one parallel to the slope.

The parallel component of gravity g can be calculated as
`\(g * \sin(\text{slope angle})\):`

`\(g = 9.81 \, \text{m/s}^2\)` (acceleration due to gravity)

Slope angle =
`\(8.44^\circ\)`

`\(g_{\text{parallel}} = 9.81 \, \text{m/s}^2 * \sin(8.44^\circ)\)`

`\(g_{\text{parallel}} \approx 9.81 \, \text{m/s}^2 * 0.147\)` (rounded to three decimal places)

`\(g_{\text{parallel}} \approx 1.442 \, \text{m/s}^2\)`

The skier is moving uphill, opposing the direction of the parallel component of gravity. To bring the skier to a stop, the acceleration he experiences will be his initial velocity
`(\(8.33 \, \text{m/s}\))` divided by the time (t) required to stop, assuming constant acceleration.

The kinematic equation used is:

`\[v = u + at\]`

Where:

v = final velocity (0 m/s, as the skier comes to a stop)

u = initial velocity (8.33 m/s)

a = acceleration
`(\(-1.442 \, \text{m/s}^2\))`

t = time taken to stop

Rearranging the equation to solve for time:

`\[t = (v - u)/(a)\]`

`\[t = (0 - 8.33)/(-1.442)\]`

`\[t \approx (-8.33)/(-1.442)\]`

`\[t \approx 5.776 \, \text{s}\]`

Question:

A skier is moving
`$8.33 \mathrm{~m} / \mathrm{s}$` when he starts to slide UP a
`$8.44^(\circ)$` frictionless slope. How much time does it take him to come to a stop?

(Unlt = s)