CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A $$2m$$ ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate $$25$$ cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is $$1$$m above the ground is?


A
253
loader
B
25
loader
C
253
loader
D
253
loader

Solution

The correct option is C $$\dfrac{25}{\sqrt{3}}$$
$$x^2+y^2=4\left(\dfrac{dy}{dt}=-25\right)$$
$$x=\dfrac{dx}{dy}+y\dfrac{dy}{dt}=0$$
$$\sqrt{3}\dfrac{dx}{dt}-1(25)=0$$
$$\dfrac{dx}{dt}=\dfrac{25}{\sqrt{3}}$$cm/sec.
1250589_1615166_ans_536010051fba4dc98c0d2b8e98801e76.png

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image