Chapter 10, Problem 42RE

To determine

To find:

a. The parametric equations that model the motion of the train and Mary as a function of time.

[Hint: The position $s$ at time $t$ of an object having acceleration $a$ is $s=\frac{1}{2}a{t}^2$.]

Answer to Problem 42RE

a. Train: $\left(\frac{3}{2}{t}^2,1\right)$; Mary: $\left(6\left(t-2\right),3\right)$.

Explanation of Solution

Given:

Mary’s train leaves at 7:15 am and accelerates at the rate of 3 meters per second. Mary, who can run 6 meters per second, arrives at the train station 2 seconds after the train has left.

Calculation:

$s=\frac{1}{2}a{t}^2$

a. Train:

$a=3m/sec$

$x_1=\frac{1}{2}*3t^2=\frac{3}{2}{t}^2$

Let $y_1=1$

Mary:

$a=6m/sec$

$x_2=6\left(t-2\right)$

Let $y_2=3$

To determine

To find:

b. Algebraically whether Mary will catch the train. If so, when?

Answer to Problem 42RE

b. No, Mary will not catch the train.

Explanation of Solution

Given:

Mary’s train leaves at 7:15 am and accelerates at the rate of 3 meters per second. Mary, who can run 6 meters per second, arrives at the train station 2 seconds after the train has left.

Calculation:

b. Mary will catch the train when $x_1=x_2$.

$\frac{3}{2}{t}^2=6\left(t-2\right)$

$t^2-4t+8=0$

$b^2-4ac=16-32<0$

$t$ does not have a real solution.

Hence, Mary will not catch the train.

To determine

To find:

c. Simulate the motions of the train and Mary by simultaneously graphing the equations found in part (a).

Answer to Problem 42RE

c. Graph is plotted.

Explanation of Solution

Given:

Mary’s train leaves at 7:15 am and accelerates at the rate of 3 meters per second. Mary, who can run 6 meters per second, arrives at the train station 2 seconds after the train has left.

Calculation:

c. Graph of equation found in part a.