Chapter 3.2, Problem 36SSC

To determine

To Calculate: The distance travelled by the in-line skater.

Answer to Problem 36SSC

  $33.74\text{m}$

Explanation of Solution

Given:

The in-line skater first accelerates from $\text{0 m/s to 5}\text{.0 m/s}$ in $4.5\text{ s}$ , and then moves at this constant speed for another $4.5\text{ s}$ .

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:

Acceleration can be calculated as follows:

  $\begin{array}{c}{v}_{final}=v_{initial}+a\Delta t\\ 5=0+a\left(4.5\right)\\ a=1.11{\text{m/s}}^{\text{2}}\end{array}$

Now, assume that the initial position of the car is $d_{initial}=0\text{m}$ , then the distance travelled in first $4.5\text{ s}$ be:

  $\begin{array}{c}{d}_1=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ d_1=0+0\left(4.5\right)+\frac{1}{2}\left(1.11\right){\left(4.5\right)}^2\\ d_1=11.24\text{m}\end{array}$

Now, the skater continues to move at the constant final velocity for another $4.5\text{ s}$ , then the distance travelled during this time is

  $\begin{array}{c}{d}_2=v_{final}\Delta t\\ =5\times 4.5\\ =22.5\text{m}\end{array}$

Then, the total distance covered by the skater is,

  $\begin{array}{c}{d}_t=d_1+d_2\\ =\left(11.24+22.5\right)\text{m}\\ =33.74\text{m}\end{array}$

Conclusion:

Thus, the total distance travelled by the skater is $33.74\text{m}$ .