The top of a 15 - foot ladder leans against a wall, and the bottom of the ladder

Question

The top of a 15 - 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 4 feet away from the wall at time t = 0. ft/sec A police car travelling south toward Sioux Falls at 160 Km/h pursues a truck travelling east away from Sioux Falls, Iowa, at 140 Km/h (See figure below). At time t = 0, the police car is 50 km north and the truck is 60 km east of Sioux Falls. Calculate the rate at which the distance between the vehicles is increasing at t = 10 min. () km/h

Explanation / Answer

7.

Let, height of top of ladder from bottom of wall be denoted by y

Distance of bottom of ladder from bottom of wall denoted by x

So,

x^2+y^2=L^2=constant

L = length of Ladder =15 foot

x^2+y^2=15^2=225

Differentiating w.r.t. gives

2xx'+2yy'=0

At t=0 x=4

Hence, at, t=1

x=4+5=9 feet

9^2+y^2=225

y^2=144

y=12 feet

x'= 5 feet/s

2xx'+2yy'=0

9*5+12y'=0

y'=-45/12=-15/4 ft/s

y'=-3.75 ft/s

8.

Denote distanct of cop car from truck as :D

D^2=x^2+y^2

t=10 min=1/6 hour
At t=0,x= 60 km,y=50 km

At t=10 min=1/6 hour,

x=60+140/6=60+70/3=250/3 km

y=50-160/6=50-80/3=70/3 km

x^2+y^2=D^2

Differentiating w.r.t. t gives

xx'+yy'=DD'

x'=140,y'=-160,x=250/3,y=70/3

D=sqrt{x^2+y^2}=10/3sqrt{25^2+7^2}=10sqrt{674}/3 km

Hence,

140*250/3-160*70/3=D'*10sqrt{674}/3

This gives

D'~91.67 km/h

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