Suppose Jos car is parked 18 meters from her house Jo is 3 m

Suppose Jo’s car is parked 18 meters from her house. Jo is 3 meters from her house, directly between her house and her car, when all of a sudden she begins walking towards her car a constant speed. 4 seconds after she started walking Jo is 9 meters from her house.

Let d represent Jo’s distance from her house (in meters) and let t represent the number of seconds since Jo started walking

a.

b

c.

Let d represent Jo\'s distance from her house (in meters) and let t represent the number of seconds since Jo started walking d=9 d=??? t=4 t=7.6 a. Complete the following sentence by filling in the answer-boxes with numbers: Jo walks at a constant speed and covers 6 meters every 4 seconds. Since 1 second is4 times as large as 4 seconds, every time Jo walks for 1 second her change in distance from the house istimes as large as 6 meters. Therefore, Jo walks 1.5meters every second. Jo walks at a constant speed and covers 6 meters every 4 seconds, Since 1 second is 4 times as large as 6 meters. Therefore, Jo walks 11.5 b. Suppose the value of t increases from 4 to 7.6. Complete the following sentence by filling in the answer-boxes with numbers: Jo walks at a constant speed and covers 6 meters in 4 seconds. Since 3.6 seconds is times as large as 4 seconds, every time Jo walks for 3.6 second her change in distance will be -times as large as 6 meters. Therefore, Jo walks meters every 3.6 seconds. c. Complete the following sentence by filling in the answer-boxes with numbers: Jo is 9 meters from her house when she has walked for 4 seconds. If Jo walks for an additional 3.6 seconds, then she will have traveled an additional after Jo started walking she is meters. Therefore, 7.6 seconds meters froum ber bouse. meters from her house

Solution

Since 1 second is (1/4) times as large as 4 seconds , every time Jo walks for 1 sec

second her change in distance from the house is ( 1/4) times as large as 6mt.

--- Reason : as she moves 6mt in 4sec so, in 1 sec she moves (1/4)*6 mt from her house which 14th of 6mt

b) Since 3.6 sec is ( 3.6/4 = 0.9) times as large as 4 seconds.

every time Jo walks for 3.6 sec her change in distance is (0.9) times as large as 6mt. ---- [ 6/4)*3.6 =6*0.9 ]

therefore Jo walks 0.9*6 = 5.4 mt

c) constant speed = (9-3)/4 = 6/4 = 1.5 mt/sec.

she will have travelled an additional 1.5*3.6 = (5.4) metres

Therefore , 7.6 second after Jo started walking she is =7.6*1.5 +3 = 14.4 mt from her house