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Question is long and as follows, I put in bold what I think is important

"A car traveling south is chasing a motorcycle that has turned at an intersection and is now traveling due east. When the car is 1 km north of the intersection and the motorcycle is 1.5 km to the east of the intersection, the distance between them is increasing at 40 km/h. What is the speed of the motorcycle, if the car is traveling at 120 km/h? Assume the two roads are straight and at right angles."

I sketched a picture of this question and it should be a Pythagoras triangle letting x= 1m and y= 1.5. You will need to find Z which is $\sqrt{3.25}$.

$z'$ is 40km/h . $y'$ is 120km/h . $x'$ is unknown

$z^2 = x^2 + y^2$

Differentiate

$2 z z' = 2 x x' + 2 y y'$

Sub in values

$2 * \sqrt{3.25} * 40 = 2 * 1.5 * x' + 2 * 120 * 1$

Now $x'$ would equal 31.93 km/h which I am skeptical about as it is so much slower than the 120km /h of the car

Is my working correct and is the motorcycle actually much slower than the car?

Feel free to correct my differentiation and square root symbols. I am not sure how to express them

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Your mistake is that $y' = -120$ km/hr. I.e. the distance between the car and the intersection is decreasing.

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