Question

A railroad track and a road cross at right angle. An observer stands on the road 70 meters south of the crossing and watches an eastbound train ntraveling at 60 meters per socond. At how many meters per second is the train moving away from the observer 4 seconds after it passess through the intersection?
A. 57.60
B. 57.88
C.59.20
D. 60.00
E. 67.40

Answer 1

Answer:

The train is moving with a speed of 57.6 m/s.    

Explanation:

Given that,

Distance of observer from the road, d = 40 m (due south)

Velocity of the train, v = 60 m/s (due east)

So,

$x(t)=70\\\\y(t)=60t$

t is time

Net displacement is given by :

$z(t)=\sqrt{(60t)^2+(70)^2}$

Differentiating above equation wrt t as :

$z'(t)=\dfrac{1}{2}(3600t^2+4900)^{-1/2}{\cdot} 7200t$

Put t = 4 s

$z'(4)=\dfrac{1}{2}(3600(4)^2+4900)^{-1/2}{\cdot} 7200(4)\\\\z'(4)=57.6\ m$

So, the train is moving with a speed of 57.6 m/s.