Bug moving on paper

Question

A bug moves linearly with constant speed across my graph paper. I first notice the bug when it is at (3, 4). It reaches (9, 8) after two seconds and (15,12) after four seconds. a) Predict the position of the bug after six seconds; after nine seconds; after t seconds. b) Is there a time when the bug is equidistant from the x- and y-axes? If so, where is it?

Answer 1

Hi Ellie,

 

If the bug is moving linearly we can use the given points to find the equation of the line

 

so

 

lets find the slope of the line = (y-y)/(x-x)

 

(8-4)/(9-3)  = 4/6 or 2/3

 

Y = (2/3)x + b

 

and we can find b by substituting one of the points

 

4 = (2/3)(3) + b

 

4 = 2 + b

 

b = 2

 

So the equation of the line is

 

y = (2/3)x + 2

 

Also from the information given we can see that the x value increases by 6 for each 2 second increase in time

 

x = 3   when t = 0

x = 9  when t = 2

x = 15 when t = 4

 

when t = 6          x = 21

 

y = (2/3)x + 2

y = (2/3)21 + 2

Y = 16

(21,16)

 

when t = 9          x = 30  

y = (2/3)x + 2
y = (2/3)30 + 2
Y = 22
(30,22)

 

after t seconds x = t/2 * 6 + 3

 

y = (2/3)(3t+3) + 2

 

y = 2t + 4

 

 

The bug would be equidistant if x = y  so lets try that

 

x = (2/3)x + 2

 

(1/3)x = 2

 

x = 6    (so at this value of x we should get a y value  of 6

 

lets try it

 

y =(2/3)x + 2

y = (2/3)6 + 2

y = 4 + 2

y=6

 

This would occur one second after the start

 

Hope this helps