A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 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)?

Answer :

Answer:

Speed along the line joining both = 7.84 meter/second  

Step-by-step explanation:

To find the rate of change of distance between the man and the car at that instant we have to find the component of speed of the car along the line joining the car and the person .

The person is 10 metre east of intersection and the car is 24 metres north of the intersection.Thus it forms a right triangle with sides of length 10 and 24.

Then the length of other side is x.

By Pythagoras theorem ,

x = [tex]\sqrt{(24)^{2}+10^{2} }[/tex]

x = 26 meters

Speed along north direction = 13 meters/second

Speed along the line joining both = [tex] 13\times cos(p)[/tex]

                                                        = [tex] 13\times cos(\frac{24}{26})[/tex]

                                                         = [tex] 13 \times 0.6034[/tex]

Speed along the line joining both = 7.84 meter/second                                                    

Answer:

-12

Step-by-step explanation:

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