Question

On the number line, a bug A is moving in the positive direction from the point 0, and a bug Bis moving in the negative direction from the point 0. The speeds of both bugs are constant. Both leave 0 at the same time, and three seconds later A has reached 18 and B has reached –15.
If x is the point reached by A after t seconds, and y is the point reached by Bafter t seconds, express x and y in terms of t.

Answer 1

Answer:

$x=st$

$y=-5t$

Step-by-step explanation:

we know that

The speed is equal to divide the distance by the time

so

The distance is equal to multiply the speed by the time

Let

s ----> the speed in units per second

d ---> the distance in units

t ---> is the time in seconds

BUG A

we have

$d=18\ units$

$t=3\ sec$

The speed is equal to

$s=\frac{d}{t}$

substitute

$s=\frac{18}{3}=6\ units/sec$

If x is the point reached by A after t seconds

then

Multiply the speed by the time

$x=st$

$x=6t$

BUG B

we have

$d=-15\ units$

$t=3\ sec$

The speed is equal to

$s=\frac{d}{t}$

substitute

$s=\frac{15}{3}=5\ units/sec$

If y is the point reached by B after t seconds

then

Multiply the speed by the time

$y=st$

$y=-5t$  ----> is negative because B is moving in the negative direction

Answer 2

Answer:

x = 6t

y = 5t

Step-by-step explanation:

We know the distance formula:

D = RT

Where

D is distance

R is speed

T is time

Bug A went 18 units after 3 seconds, so the speed is:

18 = R(3)

R = 18/3

R = 6 unit/sec

Bug B went 15 units in 3 seconds, so the speed is:

15 = R(3)

R = 15/3

R = 5 units/sec

x is reached by A after 3 seconds, so

x = 6t

y is reached by B after 3 seconds, so

y = 5t