Question

The equation d = 2t^ gives the distance from the start point of a toy boat that starts at rest and accelerates at 4 cm/s2. The equation d = 10t - 8 gives the distance from the start point of a second boat that starts at rest 8 cm behind the first boat and travels at a constant rate of 10 cm/s.

a) By setting the equations equal to each other, you can determine when the cars are the same distance from the start point: 2t2 = 10t - 8. Rewrite the equation so it is equal to 0 on the right hand side of the equation.

b) Factor the expression on the left side of the equation.

c) boats are the same distance from the start point at t = 1 and t = 4. Explain how the factors you found in part b were used to find these two times.

Answer 1

Explanation:

Given that:

`d = 2t^2` and
`d = 10t - 8`

If we equate both since they are distances, we have:

`2t^2 = 10t - 8`

Algebraically, moving the equation to the right side, we get:

`2t^ 2 = 10t -8 \\ \\ 2t^2 -10t +8 =0`

To factor the equation on the left side:

`2t^2 - 10t + 8 = 0`

Factor out 2 on the left-hand side:

`2 (t^2 -5t + 4) =0`

`=2 (t^2 -t-4t + 4)`

`= 2[t(t-1)-4(t-1)]`

`=2(t-1) (t-4)`

To determine the time when the two boats have to cover an equal distance:

`2(t -1)(t-4) =0 \\ \\(t-1)(t-4) =0`

Thus:

`t - 1 = 0 \ or \ t - 4 = 0 \\ \\ t = 1 \ or \ t = 4`