Question 1163169
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Let t=0 (hours) be at 2pm.  Then<br>
The position of train A at t=0 is (120,0). (At 2pm, it has been traveling east at 60km/h.)
The position of train A at time t is (60t+120,0).<br>
The position of train B at time t is (0,75t).<br>
The distance between them at time t is<br>
{{{d = sqrt((60t+120)^2+(75t)^2)}}}<br>
We need to find the rate at which that distance is increasing at 5pm (at t=3).<br>
To avoid differentiating an ugly square root expression, I will use the square of the distance....<br>
{{{d^2 = (60t+120)^2+(75t)^2 = 3600t^2+14400t+14400+5625t^2 = 9225t^2+14400t+14400}}}<br>
{{{2d*(dd/dt) = 18450t+14400}}}<br>
We need to evaluate dd/dt at t=3; that means we need to know the distance d between the trains at t=3:<br>
{{{d(3) = sqrt(300^2+225^2) = 375}}}<br>
(Note that calculation is easy if you recognize that the legs 225 and 300 are scale models of a 3-4-5 right triangle, making the hypotenuse 375.)<br>
So at t=3,<br>
{{{2d*(dd/dt) = 18450t+14400}}}<br>
{{{2(375)*(dd/dt) = 18450(3)+14400 = 69750}}}<br>
{{{dd/dt = 69750/750 = 93}}}<br>
ANSWER: At 5pm, the two trains are separating at a rate of 93km/h.<br>