Question

A computer animation below shows a cat moving in a straight line:-

Its height, b meters, above the ground, is given by 8S - 3b = -9, where s is the time in
seconds after it stars moving. In the same animation, a mouse starts to move at the same
time as the cat and its movement is given by -3s + b = 1,

1) Will the mouse be able to catch the cat ?
2) If Yes, after how much time and at what height?

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Answer 1

Answer:

Therefore the mouse will catch the cat after 6 seconds at 19 meters height.

Step-by-step explanation:

The mouse will be able to catch the cat in the time, s, in seconds if they both move at the same height.

for the cat we get                (8s + 9) / 3 = b

for the mouse we get          1 +3s     = b

therefore equating the two we get

 (8s + 9) / 3 =  1 +3s       therefore 8s + 9 = 3 + 9s

therefore s = 6 seconds, and b = 3×6 + 1 = 19meters

Therefore the mouse will catch the cat after 6 seconds at 19 meters height.

Answer 2

Answer:

A computer animation below shows a cat moving in a straight line:-

Step-by-step explanation:

Its height, b meters, above the ground, is given by 8S - 3b = -9, where s is the time in

seconds after it stars moving. In the same animation, a mouse starts to move at the same

time as the cat and its movement is given by -3s + b = 1,

1) Will the mouse be able to catch the cat ?

2) If Yes, after how much time and at what height?