Chapter 2, Problem 88P

(a)

To determine

Minimum negative acceleration to avoid collision.

Answer to Problem 88P

Minimum negative acceleration to avoid collision is $0.75\text{m}/{\text{s}}^{\text{2}}$ .

Explanation of Solution

Given:

Velocity of passenger train is $29\text{m}/\text{s}$ .

Separation between passenger and freight train when engineer sees it is $360\text{m}$ .

Velocity of freight train is $6\text{m}/\text{s}$ .

Reaction time of engineer is $0.4\text{s}$

Formula used:

Write the expression for relative velocity.

  $v_{ir}=(v_1-v_{{}_2})$  ........(1)

Here, $v_{ir}$ is initial relative velocity, $v_1$ is velocity of passenger train and $v_2$ is velocity of freight train

Write the expression for distance covered between them during reaction time .

  $s=(v_1-v_2)t_r$  ........(2)

Here, $s$ is distance covered between them during reaction time and $t_r$ is reaction time

Write the expression for remaining distance between trainsafter reaction time

  $s_1=(360-s)$  ........(3)

Here, $s_1$ is remaining distance between trains after reaction time.

Write the expression for final relative velocity between the trains

  $v_{rf}{}^2=(v_{ri}{}^2-2a_{\min }{s}_1)$  ........(4)

Here, $v_{rf}$ is final relative velocity which will be equal to zero in the case.

Calculation:

Substitute $29\text{m}/\text{s}$ for $v_1$ and $6\text{m}/\text{s}$ for $v_2$ in equation (1).

  $\begin{array}{l}{v}_{ir}=(29-6)\\ v_{ir}=23\text{m}/\text{s}\end{array}$ .

Substitute $29\text{m}/\text{s}$ for $v_1$ , $6\text{m}/\text{s}$ for $v_2$ and $0.4\text{s}$ for $t_r$ in equation (2).

  $\begin{array}{l}s=(29-6)0.4\\ s=9.2\text{m}\end{array}$ .

Substitute $9.2\text{m}$ for $s$ in equation (3)

  $\begin{array}{l}{s}_1=(360-9.2)\\ s_1=350.8\text{m}\end{array}$

Substitute $23\text{m}/\text{s}$ for $v_{ir}$ and $350.8\text{m}$ for $s_1$ in equation (4)

  $\begin{array}{c}0={23}^2-2(a_{\min })(350.8)\\ a_{\min }=0.75\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the minimum negative acceleration to avoid collision is $0.75\text{m}/{\text{s}}^{\text{2}}$

(b)

To determine

Velocity of passenger train at minimum acceleration to avoid collision

Answer to Problem 88P

Velocity of passenger train at minimum acceleration to avoid collision is $10.07\text{m}/\text{s}$ .

Explanation of Solution

Given:

Reaction time is $0.8\text{s}$

Formula used:

Write expression for distance remaining between trains after reaction time

  $s_2=360-(v_1-v_2)t_r$  ........(5)

Here, $s_2$ is distance remaining between trains after reaction time and $t_r$ is reaction time.

Write expression for final relative velocity.

  $v_{fr}=\sqrt{v_{ir}-2a{s}_2}$  ........(6)

Here, $v_{fr}$ is final relative velocity.

Write expression for final relative velocity.

  $v_{fr}=v_1-v_2$  ........(7)

Calculation:

Substitute $29\text{m}/\text{s}$ for $v_1$ and $6\text{m}/\text{s}$ for $v_2$ and $0.8\text{s}$ for $t_r$ in equation (5)

  $\begin{array}{l}{s}_2=360-(23)0.8\\ s_2=341.6\text{m}\end{array}$ .

Substitute $23\text{m}/\text{s}$ for $v_{ir}$ , $0.75\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $341.6\text{m}$ for $s_2$ in equation (6)

  $\begin{array}{l}{v}_{fr}=\sqrt{{23}^2-2(0.75)(341.6)}\\ v_{fr}=4.07\text{m}/\text{s}\end{array}$ .

Substitute $4.07\text{m}/\text{s}$ for $v_{fr}$ and $6\text{m}/\text{s}$ for $v_2$ in equation (7)

  $\begin{array}{l}4.07=v_1-6\\ v_1=10.07\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity of passenger train at minimum acceleration to avoid collision is $10.07\text{m}/\text{s}$ .

(c)

To determine

Distance covered by passenger train between sighting and collision

Answer to Problem 88P

The distance covered by passenger train between sighting and collision is $516.16\text{m}$ .

Explanation of Solution

Formula used:

Write the expression for time when train just collides.

  $s_2=v_{ir}t-\frac{1}{2}\left(0.75t^2\right)$  ........(8)

Write the expression for distance covered by passenger train between sighting and collision

  $s_3=v_1(t+t_r)-\frac{1}{2}a{t}^2$  ........(9)

Here, $s_3$ is distance covered by passenger train between sighting and collision, $t$ is time when train just collides.

Calculation:

Substitute $341.6\text{m}$ for $s_2$ and $23\text{m}/\text{s}$ for $v_{ir}$ in equation (8)

  $\begin{array}{c}341.6=23t-\frac{1}{2}\left(0.75t^2\right)\\ t=25.23\text{s}\end{array}$ .

Substitute $29\text{m}/\text{s}$ for $v_1$ , $25.23\text{s}$ for $t$ and $0.8\text{s}$ for $t_r$ in equation (9).

  $\begin{array}{l}{s}_3=29(25.23+0.8)-\frac{1}{2}(0.75){(}^{25.23}\\ s_3=516.16\text{m}\end{array}$

Conclusion:

Distance covered by passenger train between sighting and collision is $516.16\text{m}$ .