Question

Lisa is standing on a dock beside a lake. She drops a rock from her hand into the lake. After the rock hits the surface of the lake, the rock's distance from the lake's surface changes at a rate of −2 inches per second. If Lisa holds her hand 5 feet above the lake's surface, how far from Lisa's hand is the rock 4 seconds after it hits the surface?

Answer 1

Distance after 8s:-

`\\ \rm\longmapsto 4(-2)=-8in`

Now

Distance:-

`\\ \rm\longmapsto 5-(-8)`

`\\ \rm\longmapsto 5ft+8in`

• 1ft=12in

`\\ \rm\longmapsto 5ft+3/4ft`

`\\ \rm\longmapsto 23/4ft`

`\\ \rm\longmapsto 5.7ft`

• Convert to in

`\\ \rm\longmapsto 5.7* 12`

`\\ \rm\longmapsto 68in`

Answer 2

• 5 ft 8 in or 68 in

Explanation:

In 4 seconds the rock will be at:

• 2*4 = 8 in from the surface

Add 5 feet to this to get the required distance:

• 5 ft + 8 in = 5*12 in + 8 in = 68 in