Question

1. a railroad track and a road cross at right angles. an observer stands on the road 70 meters south of the crossing and watches an east - bound train traveling at 60 meters per second. at how many meters per second is the train moving away from the observer 4 seconds after it passes through the intersection? (a) 57.60 (b) 57.88 (c) 59.20 (d) 60.00 (e) 67.40

Explanation:

Step1: Define variables

Let $x$ be the distance of the train from the intersection and $y = 70$ (constant distance of the observer from the intersection), and $z$ be the distance between the train and the observer. By the Pythagorean theorem, $z^{2}=x^{2}+y^{2}$. Since $y = 70$, we have $z^{2}=x^{2}+4900$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of $z^{2}=x^{2}+4900$ with respect to $t$ gives $2z\frac{dz}{dt}=2x\frac{dx}{dt}$. Then $\frac{dz}{dt}=\frac{x}{z}\cdot\frac{dx}{dt}$.

Step3: Find $x$ and $z$ at $t = 4$ seconds

The train is moving at a speed of $\frac{dx}{dt}=60$ m/s. After $t = 4$ seconds, $x=60\times4 = 240$ meters. Using the Pythagorean - theorem, $z=\sqrt{x^{2}+y^{2}}=\sqrt{240^{2}+70^{2}}=\sqrt{57600 + 4900}=\sqrt{62500}=250$ meters.

Step4: Calculate $\frac{dz}{dt}$

Substitute $x = 240$, $z = 250$, and $\frac{dx}{dt}=60$ into $\frac{dz}{dt}=\frac{x}{z}\cdot\frac{dx}{dt}$. We get $\frac{dz}{dt}=\frac{240}{250}\times60=\frac{14400}{250}=57.6$ m/s.

Answer:

A. 57.60