The top of a 20-foot ladder leans against a wall, and the bottom of the ladder slides away from the wall at a rate of 5 ft/sec. find the velocity of the top of the ladder at time t=1 sec if the bottom of the ladder is 7 feet away from the wall at time t=0

vy= -3.75 ft/sec (slides down )

Explanation:

denoting y as the coordinate of the top of the ladder , x as the coordinate of the bottom , L as the length of the ladder , then

x²+y²=L² = constant

deriving by t

2*x*dx/dt + 2*y*dy/dt = 0

since the velocity is vx=dx/dt and vy=dy/dt

x*vx+y*vy=0

then since the velocity of x is constant x=x₀+vx*t

at t=1 sec x=7 ft +  5 ft/sec*1 sec = 12 ft

then x²+y²=L² → y= √(L²-x²) =√[(20 ft)²-(12 ft)²] = 16 ft

after that

x*vx+y*vy=0 → vy=  -(x/y)*vx = -(12 ft/16 ft)*5 ft/sec= -3.75 ft/sec

vy= -3.75 ft/sec (slides down since vx is positive, that means that slides away)

Note

we cannot assume that vy is constant ( and do y=y₀+vy*t)  when vx is constant , since y= √(L²-x²) = √(L²-(x₀+vx*t)²) , that is not linear with t . Nevertheless, vy could be found deriving y with respect to t , but that would be more complicated than the other way.

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when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min