Chapter 2, Problem 63GP

To determine

The speed of the last car as it passes the worker

Answer to Problem 63GP

  $30.8\frac{\text{m}}{\text{s}}$

Given:

For front car of the train:

Initial velocity of the front car as the train start $=v_{fo}=0\frac{\text{m}}{\text{s}}$

Final velocity of the front car as it reaches the worker $=v_{ff}=25\frac{\text{m}}{\text{s}}$

Initial position of front car as the train starts $=x_{fo}=95\text{m}$

Final position of front car as it reaches the worker $=x_{ff}=95+180=275\text{m}$

Acceleration of the train $=a$

For last car of the train when front car is at the location of the worker:

Initial velocity of the last car as front car passes the worker $=v_{lo}=25\frac{\text{m}}{\text{s}}$

Final velocity of the last car as it passes the worker $=v_{lf}$

Initial position of last car as front car passes the worker $=x_{lo}=180\text{m}$

Final position of last car as it reaches the worker $=x_{ff}=95+180=275\text{m}$

Acceleration of the train $=a$

Formula Used:

Final position during accelerated motion of an object is given as

  $x_{ff}=x_{fo}+\frac{v_f^2-v_o^2}{2a}$

Calculation:

Consider the motion of the front car:

Final position of the front car at the position of worker is given as

  $x_{ff}=x_{fo}+\frac{v_{ff}^2-v_{fo}^2}{2a}$

Substituting the values

  $\begin{array}{l}275=95+\frac{{25}^2-0^2}{2a}\\ 275-95=\frac{625}{2a}\\ a=\frac{625}{2(180)}\\ a=1.7\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

Consider the motion of the last car:

Final position of the last car at the position of worker is given as

  $x_{lf}=x_{lo}+\frac{v_{lf}^2-v_{lo}^2}{2a}$

Substituting the values

  $\begin{array}{l}275=180+\frac{v_{lf}^2-{25}^2}{2(1.7)}\\ 275-180=\frac{v_{lf}^2-625}{3.4}\\ v_{lf}^2-625=(95)(3.4)\\ v_{lf}=30.8\frac{\text{m}}{\text{s}}\end{array}$

Conclusion:

Therefore, speed of last car is $30.8\frac{\text{m}}{\text{s}}$

Explanation of Solution

Given:

For front car of the train:

Initial velocity of the front car as the train start $=v_{fo}=0\frac{\text{m}}{\text{s}}$

Final velocity of the front car as it reaches the worker $=v_{ff}=25\frac{\text{m}}{\text{s}}$

Initial position of front car as the train starts $=x_{fo}=95\text{m}$

Final position of front car as it reaches the worker $=x_{ff}=95+180=275\text{m}$

Acceleration of the train $=a$

For last car of the train when front car is at the location of the worker:

Initial velocity of the last car as front car passes the worker $=v_{lo}=25\frac{\text{m}}{\text{s}}$

Final velocity of the last car as it passes the worker $=v_{lf}$

Initial position of last car as front car passes the worker $=x_{lo}=180\text{m}$

Final position of last car as it reaches the worker $=x_{ff}=95+180=275\text{m}$

Acceleration of the train $=a$

Formula Used:

Final position during accelerated motion of an object is given as

  $x_{ff}=x_{fo}+\frac{v_f^2-v_o^2}{2a}$

Calculation:

Consider the motion of the front car:

Final position of the front car at the position of worker is given as

  $x_{ff}=x_{fo}+\frac{v_{ff}^2-v_{fo}^2}{2a}$

Substituting the values

  $\begin{array}{l}275=95+\frac{{25}^2-0^2}{2a}\\ 275-95=\frac{625}{2a}\\ a=\frac{625}{2(180)}\\ a=1.7\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

Consider the motion of the last car:

Final position of the last car at the position of worker is given as

  $x_{lf}=x_{lo}+\frac{v_{lf}^2-v_{lo}^2}{2a}$

Substituting the values

  $\begin{array}{l}275=180+\frac{v_{lf}^2-{25}^2}{2(1.7)}\\ 275-180=\frac{v_{lf}^2-625}{3.4}\\ v_{lf}^2-625=(95)(3.4)\\ v_{lf}=30.8\frac{\text{m}}{\text{s}}\end{array}$

Conclusion:

Therefore, speed of last car is $30.8\frac{\text{m}}{\text{s}}$