A popular web video shows a jet airplane, a car, and a motorcycle racing from rest along a runway (Fig 2-10 ). Initially the motorcycle takes the lead, but then the jet takes the lead, and finally the car blows past the motorcycle. Here let's focus on the car and motorcycle and assign some reasonable values to the motion. The motorcycle first takes the lead because its ( constant) acceleration `a_(m)=8.40 m//s^(2)` is greater than the car's ( constant) acceleration `a_(c) = 5.60 m//s^(2)`, but it soon loses to the car because it reaches its greater speed `v_(m)= 58.8m//s` before the car reaches the greatest speed `v_(c) = 106 m//s.` How long does the car take to reach the motorcycle ?

Explanation

KEY IDEA
We can apply the equations of constant acceleration to both vehicles, but for the motorcycle we must consider the motion in two stages: (1) First it travels through distance `x_(m l)` with zero initial velocity and acceleration `a_(m)=8.40m//s^(2)`, reaching speed `v_(m)=58.8m//s.` (2) Then it travels through distance `x_(m2)` with constant velocity `v_(m)= 58.8m//s` and zero acceleraton ( that, too, is a constant acceleraton). Note that we symbolized the distance even though we do not know their values, Symbolizing unknown quantities is often helpful in solving physics problems.)
Calculations: So that we can draw figures and do caculatons, let.s assume that the vehicles race along he positive direction of an x axis, starting from `x=0` at time `t=0`. (We can choose any initial numbers because we are looking for the elapsed time, not a particular time in, say, the afternoon, but let.s stick with these easy numbers.) We

want the car to pass the motorcycle, but what does that mean mathematically ?
It means that at some time t, the side-by-side vehicles are at the same coordinate: `x_(c)` for the car and the sum `x_(m1)+x_(m2)` for the motorcycle. We can write this statement mathematically as
`x_(c)= x_(m1)=x_(m2)`, (2-19)
Now let.s fill out both sides of Eq. 2-19, left side first. To reach the passing point at `x_(c)`, the car accelerates from rest . From Eq. 2-15 `(x-x_(0)= v_(0)t+1/2 a t^(2))`, with `x_(0)` and `v_(0)=0`, we have
`x_(c)= 1/2 a_(c)t^(2)` (2-20)
To write an expression for `x_(m1)` for the motorcycle, we first find the time `t_(m)` it takes to reach its maximum speed `v_(m)`, using Eq. 2-11 `(v=v_(0)+ a t)`. Substituting `v_(0)=0, v= v_(m)=58.8m//s`, and `a=a_(m)=8.40 m//s^(2)`, that time is
`t_(m)= (v_(m))/(a_(m))= (58.8 m//s)/(8.40 m//s^(2))=7.00s`. (2-21)
To get the distance `x_(m1)` traveled by the motorcycle during the first stage, we again use Eq. 2-15 with `x_(0)=0` and `v_(0)=0`, but we also substitute from Eq. 2-21 for the time. We find
`x_(m1)= 1/2 a_(m) t_(m)^(2) = 1/2 a_(m) ((v_(m))/(a_(m)))^(2)=1/2 (v_(m)^(2))/(a_(m))`. (2-22)
For the remaining time of `t-t_(m)`, the motorcycle travels at its maximum speed with zero acceleration. To get the distance, we use Eq. 2-15 for this second stage of the motion, but now the initial velocity is `v_(0)=v_(m)` ( the speed at the end of the first stage) and the acceleration is `a=0`. So, the distance traveled during the second stage is
`x_(m2) = v_(m)(t-t_(m))= v_(m)(t-7.00s)`. (2-23)
To finish the calculation, we substitute Eqs. 2-20, 2-22, and 2-23 into Eq. 2-19, obtaining
`1/2 a_(c) t^(2)= 1/2 (v_(m)^(2))/(a_(m)) + v_(m) (t-7.00s)` (2-24)
This is a quadratic equation. Substituting in the given data, we solve the equation ( by using the usual quadratic-equation formula or a polynomial sovler on a calculator), finding `t=44.4s` and `t=16.6s`.
But what do we do with two answers? Does the car pass the motorcycle twice? No, of course not, as we can see in the video. So, one of the answers is mathematically correct but not physically meaningful. Because we know that the car passes the motorcycle after the motorcycle reaches its maximum speed at `t=7.00`s, we discard the solution with `t lt 7.00s` as being the unphysical answer and conclude that the passing occurs at
`t=16.6s`.
Figure 2-11 is a graph of the position versus time for the two vehicles, with the passing point marked. Notice that at `t=7.00`s the plot for the motorcycle switches from being curved ( because the speed had been increasing ) to being straight ( because the speed is thereafter constant).