Question

You begin sliding down a 15 ski slope. Ignoring friction and air resistance, how fast will you be moving after 10?

 

Answer 1

 

Concepts and reason

The concepts required to solve the given question are the acceleration of an object along the inclined plane and the kinematics equations, which explain the velocity, acceleration, and time. First, calculate the acceleration of the person. It depends upon the acceleration due to gravity and the slope of the plane. Then use the equation of kinematics to find the speed of the person. Finally, substitute the given values in the equation for velocity.

Fundamentals

Acceleration is the "rate of change of velocity". It is a vector quantity. In this case, the motion is along the slope. Hence the acceleration during the motion depends upon the acceleration due to gravity and the angle of inclination. Acceleration due to gravity is the "acceleration gained by an object from the gravitational force". This is experienced by all bodies that are moving under gravity. The speed of the object can be calculated using the kinematics equations. These equations can be applied to all bodies in motion having either constant velocity or constant acceleration. It gives the relation between the variables: displacement, velocity, acceleration, and time. The expression for the acceleration of the person, \(a=g \sin \theta\)

 

Here, \(a\) is the person's acceleration, \(g\) is the acceleration due to gravity, and \(\boldsymbol{\theta}\) is the angle of inclination. The equation of kinematics is, \(v=u+a t\)

 

Here, \(v\) is the final speed \(u\) is the initial speed, and \(t\) is the time taken.

 

Calculate the acceleration of the person on the ski slope, \(a=g \sin \theta\)

Acceleration of the person on the ski slope, \(a=g \sin \theta\)

Substitute \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) for \(g\) and \(15^{\circ}\) for \(\theta\) in the above equation. \(a=9.8 \mathrm{~m} / \mathrm{s}^{2} \times \sin 15^{\circ}\)

\(=2.536 \mathrm{~m} / \mathrm{s}^{2}\)

The person moving on the ski slope experiences acceleration during his motion. It depends upon the acceleration due to gravity and the angle of inclination. Acceleration of the person, \(a=g \sin \theta\)

Here, \(a\) is the person's acceleration, \(g\) is the acceleration due to gravity, and \(\boldsymbol{\theta}\) is the angle of inclination. [Hint for the next step] Calculate the speed of the person after 10 seconds from the equation of kinematics. \(v=u+a t\)

The kinematics equation of motion,

\(v=u+a t\)

Substitute \(0 \mathrm{~m} / \mathrm{s}\) for \(\mathrm{u}, 2.53 \mathrm{~m} / \mathrm{s}^{2}\) for \(\mathrm{a}\), and \(10 \mathrm{~s}\) for \(\mathrm{t}\) and solve for \(\mathrm{v}\). \(v=(0 \mathrm{~m} / \mathrm{s})+\left(2.536 \mathrm{~m} / \mathrm{s}^{2}\right) \times(10 \mathrm{~s})\)

\(=25.36 \mathrm{~m} / \mathrm{s}\)

Here, the initial velocity is zero, and also there is no friction and air resistance. Hence there is no frictional force acting in this case. Thus the speed after 10 seconds depends upon the acceleration and the time taken by the person. [Common mistake]

Acceleration of the person is equal to \(g\) times \(\sin \theta\). It is \(\operatorname{not} \cos \theta\). This can also be written as,

\(a=g \cos \left(90^{\circ}-\theta\right)\)

\(=g \cos \left(90^{\circ}-15^{\circ}\right)\)

\(=g \cos 75^{\circ}\)

The speed of the person after 10 seconds is \(25.36 \mathrm{~m} / \mathrm{s}\).