A person stands 101010 meters east of an intersection and watches a car driving towards the intersection from the north at 131313 meters per second. At a certain instant, the car is 242424 meters from the intersection. What is the rate of change of the distance between the car and the person at that instant (in meters per second)?.

We have given that a person is standing 10 meters east from junction or intersection point. Now, we have given that rate of change of car from north direction as 13m/sec.

So, let assume distance of intersection point from the final point is y meters and distance of a person from final point as x meters,

Therefore, using pythagoras theorem,we get

=>y² = x² +(10)²

=>y=√(x²+100)

Now we are given rate of change of x in form of dx/dt,so on differentiating we get

=>dy/ dt = [(1/2) × (1/√(x²+100))] × (2 × x × dx/dt)

=>dy / dt =( x /√(x²+100)) × dx/ dt

Now, we are given value of dx/dt= -13m/sec. Since, distance is decreasing with time therefore dx/dt = -13m/sec.

Now, putting value of dx/dt and x=24meters,we get

=>dy /dt = (24 / √[(24)²+100]) × (-13)

=>dy /dt = (24 / √(576+100)) × (-13)

=>dy /dt = (24/26)×(-13)

=>dy/dt = -24/2

=>dy/dt = -12m/sec

Hence, rate of change of the distance is 12m/sec.

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