Question

Physics acceleration and velocity homework question?

A brown road runner is standing still pecking at soil looking for worms.  Suddenly, a black road runner travelling at a constant speed of 9.25 m/s, overtakes the brown roadrunner.   The brown road runner notices that the other one has a worm in its mouth and begins to accelerate at 2.75 m/s2 to catch up to and steal the worm.  How far will the brown road runner have to travel to catch up to the black road runner?   Assume the stationary brown road runner begins to move the moment the black road runner passes it.

Answer 1

We need to write two kinematic equations one for brown road runner and another one for black road runner.

Brown road runner.

We note that brown road runner is standing still. It accelerate at a rate `=2.75 ms^-2` and attempts to catch up with the black road runner.
Let us assume that it is able to catch up the other road runner after time `=t`
Distance traveled by brown road runner can be found by the kinematic expression
`s=ut+1/2at^2`
Inserting given numbers we get
`s=0xxt+1/2xx2.75t^2`
`=>s=1.375t^2` ......(2)

Black road runner.

We see that this road runner travels at a constant speed `=9.25ms^-1`
This road runner also travels distance `s` in time `t` when brown catches up to black road runner.
Kinematic expression for this road runner is
`"Distance traveled"="Speed"xx"Time"`
`s=9.25t` .....(2)
To find distance `s` we eliminate `t` from equation (1) with the help of (2). From (2)
`t=s/9.25`
Inserting this value in (1) we get
`s=1.375(s/9.25)^2`
`=>1.375/(9.25)^2s^2-s=0`
`=>s(1.375/(9.25)^2s-1)=0`
Two roots of this quadratic are
`s=0`
`1.375/(9.25)^2s-1=0`

Ignoring first root as it represents when black road runner first time overtook brown road runner. We get
`s=1xx(9.25)^2/1.375`
`s=62.23m`, rounded to two decimal places.