Chapter 3.3, Problem 49SSC

To determine

To Calculate:The final velocity of the ride and the distance covered by the people on the ride.

Answer to Problem 49SSC

  $v_f=22.54\text{m/s}$ and $d_f=25.921\text{m}$

Explanation of Solution

Given:

A free-fall ride starts at rest and is in free fall.

Time elapsed: $\Delta t=2.3\text{s}$

Formula used:

Final velocity with constant acceleration can be obtained by:

  $v_{final}=v_{initial}+a\Delta t$

Where, $\Delta t$ is the time taken, a is acceleration and v represents velocity.

The distance traveled by an object with a constant acceleration can be obtained by:

  $d_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a\Delta t^2$

Here, d represents displacement, v represents velocity, t represents time and a is the acceleration.

Calculation:

Since, the ride starts from the rest, thus its initial velocity will be 0, that is

  $v_i=0\text{m/s}$

Acceleration acting on it will be due to gravity only. Thus,

  $a=9.8{\text{m/s}}^{\text{2}}$

Now, using the above-mentioned formula for the velocity, the final velocity of the ride after 2.3s will be,

  $\begin{array}{c}{v}_f=v_i+a\Delta t\\ =0+\left(9.8\right)\left(2.3\right)\\ =22.54\text{m/s}\end{array}$

The distance covered by the ride in 2.3s will be:

  $\begin{array}{c}{d}_f=d_i+v_i\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ =0+0\left(\Delta t\right)+\frac{1}{2}\left(9.8\right){\left(2.3\right)}^2\\ =25.921\text{m}\end{array}$

Conclusion:

Thus, the final velocity and the distance covered by the ride in the given time is $22.54\text{m/s}$ and $25.921\text{m}$ .