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Sagot :
The rate of change of the distance between the car and the person at that instant is 12m/sec.
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.
To know more about distance, visit here:
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