Question

A roadrunner at rest suddenly spots a rattlesnake slithering directly away at a constant speed of 0.75 m/s. At the moment the snake is 10. Meters from the bird, the roadrunner starts chasing it with a constant acceleration of 1.0 m/s^2. How long will it take the roadrunner to catch up to the snake?

Answer 1

The roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

Step-by-step explanation:

From the statement we notice that:

1) Rattlesnake moves a constant speed (
`v_(S) = 0.75\,(m)/(s)`), whereas the roadrunner accelerates uniformly from rest. (
`v_(o, R) = 0\,(m)/(s)`,
`a = 1\,(m)/(s^(2))`)

2) Initial distance between the roadrunner and rattlesnake is 10 meters. (
`x_(o, R) = 0\,m`,
`x_(o,S) = 10\,m`)

3) The roadrunner catches up to the snake at the end. (
`x_(S) = x_(R)`)

Now we construct kinematic expression for each animal:

Rattlesnake

`x_(S) = x_(o,S)+v_(S)\cdot t`

Where:

`x_(o, S)` - Initial position of the rattlesnake, measured in meters.

`x_(S)` - Final position of the rattlesnake, measured in meters.

`v_(S)` - Speed of the rattlesnake, measured in meters per second.

`t` - Time, measured in seconds.

Roadrunner

`x_(R) = x_(o,R) +v_(o,R)\cdot t +(1)/(2)\cdot a\cdot t^(2)`

Where:

`x_(o, R)` - Initial position of the roadrunner, measured in meters.

`x_(R)` - Final position of the roadrunner, measured in meters.

`v_(o,R)` - Initial speed of the roadrunner, measured in meters per second.

`a` - Acceleration of the roadrunner, measured in meters per square second.

`t` - Time, measured in seconds.

By eliminating the final positions of both creatures, we get the resulting quadratic function:

`x_(o,S)+v_(S)\cdot t = x_(o,R)+v_(o,R)\cdot t +(1)/(2)\cdot a \cdot t^(2)`

`(1)/(2)\cdot a \cdot t^(2) + (v_(o,R)-v_(S))\cdot t + (x_(o,R)-x_(o,S)) = 0`

If we know that
`a = 1\,(m)/(s^(2))`,
`v_(o, R) = 0\,(m)/(s)`,
`v_(S) = 0.75\,(m)/(s)`,
`x_(o, R) = 0\,m` and
`x_(o,S) = 10\,m`, the resulting expression is:

`0.5\cdot t^(2)-0.75\cdot t -10=0`

We can find its root via Quadratic Formula:

`t_(1,2) = \frac{-(-0.75)\pm \sqrt{(-0.75)^(2)-4\cdot (0.5)\cdot (-10)}}{2\cdot (0.5)}`

`t_(1,2) = (3)/(4)\pm (√(329))/(4)`

Roots are
`t_(1) \approx 5.285\,s` and
`t_(2)\approx -3.785\,s`, respectively. Both are valid mathematically, but only the first one is valid physically. Hence, the roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.