Chapter 3, Problem 68GP

(a)

To determine

To calculate: The skier’s vertical component of acceleration.

Answer to Problem 68GP

The vertical component of acceleration is $0.9{\text{ m/s}}^2$ and is directed downwards.

Explanation of Solution

Given data:

Acceleration, $a=1.80{\text{ m/s}}^2$

The angle of slope of the hill with the horizontal, $\theta ={30}^{\text{o}}$

Formula used:

The magnitude of vertical component of acceleration a is:

  $a_y=a\sin \theta$

Calculation:

Consider the downward direction to be positive y direction.

The acceleration of the skier, $a=1.80{\text{ m/s}}^2$

The angle of slope of the hill with the horizontal, $\theta ={30}^{\text{o}}$

The vertical component of acceleration is directed downwards, and its magnitude is given by,

  $\begin{array}{c}{a}_y=a\sin \theta \\ =(1.80{\text{ m/s}}^2{\text{)(sin 30}}^{\text{o}})\\ =0.9{\text{ m/s}}^2{\text{ [sin 30}}^{\text{o}}=0.5]\end{array}$

Conclusion:

The vertical component of acceleration is $0.9{\text{ m/s}}^2$ and is directed downwards.

(b)

To determine

To calculate: The time taken by the skier to reach the bottom of the hill.

Answer to Problem 68GP

The time taken by the skier to reach the bottom of the hill is $27.28\text{ s}$

Explanation of Solution

Given data:

Acceleration, $a=1.80{\text{ m/s}}^2$

The angle of slope of the hill with the horizontal, $\theta ={30}^{\text{o}}$

Initial velocity, $u=0\text{ m/s}$ (at rest)

Total displacement, $d=335\text{ m}$

Vertical acceleration, $a_y=0.9{\text{ m/s}}^2$

Formula used:

From laws of motion,

  $d=ut+\frac{1}{2}a{t}^2$

Where d is the total displacement of a body, u is initial velocity , t is total time of motion, and a is the acceleration of the body.

Calculation:

Consider the downward direction to be positive y direction.

Initial velocity, $u=0\text{ m/s}$ (at rest)

Total displacement, $d=335\text{ m}$

Vertical acceleration, $a_y=0.9{\text{ m/s}}^2$

Therefore, from laws of motion,

  $\begin{array}{c}d=ut+\frac{1}{2}{a}_y{t}^2\text{ [as displacement is downwards, }a=a_y\text{]}\\ 335\text{ m}=(0\text{ m/s)(}t)+\frac{1}{2}(0.9{\text{ m/s}}^2)t^2\\ 335\text{ m}=(0.45{\text{ m/s}}^2)t^2\\ t^2=\frac{335\text{ m}}{0.45{\text{ m/s}}^2}=744.44{\text{ s}}^2\\ t=27.28\text{ s}\end{array}$

Conclusion:

The time taken by the skier to reach the bottom of the hill is $27.28\text{ s}$