Question

How to find the maximum velocity of two objects that travel from a point `A` to a stopping point `B` in the same amount of time?

A train passes through a station `A` and travels for `10\ km` at a constant speed of `60\ \frac{km}{h}`, then it is uniformly decelerated for `3\ km` until stopping in station `B`. A second train leaves from the same station `A` simultaneously with the passage of the first and travels before uniformly accelerated motion and then of uniformly decelerated motion until stopping at station `B` at the same instant of the first train. What is the maximum speed reached by the two trains?

Answer 1

Train 1's maximum velocity = `60 (km)/h`.
Train 2's maximum velocity = `97.5 (km)/h`.

Explanation:

Train 1's maximum velocity is obviously `60 (km)/h`.

Train 2 will not be as easy. In fact we need to study train 1's data to get useful information to find train 2's maximum velocity. The time is the same for both, so getting train 1's time will help.

Train 1 went 10 km at `60 (km)/h` in time, `t_1`, given by

`t_1 = (10 cancel(km))/(60 cancel(km)/h) = 1/6 h`

Then train 1 put on the brakes, decelerating uniformly to a stop in 3 km. What was the acceleration, `a`, and the time, `t_2`? Because the acceleration is uniform, we can use the kinematic formula

`v^2 = u^2 + 2*a*x`

`0^2 = ((60 km)/h)^2 + 2*a*3 km`

`2*a*3 km = -((60 km)/h)^2`

`a = -((60 km)/h)^2/(2*3 km) = -600 ((km*cancel(km))/h^2)/cancel(km) = -600 (km)/h^2`

It is unusual to see an acceleration with `h^2` instead of `s^2`, but it is the proper dimensions and it will turn out OK in the end.

For the time to come to a stop, use the kinematic formula

`v = u + a*t`

`0 = 60 (km)/h - 600 (km)/h^2*t`

`600 (km)/h^2*t = 60 (km)/h`

`t = (60 cancel(km)/cancelh)/(600 cancel(km)/(h*cancel(h))) = 1/10 h`

OK, total time for either train is `(1/6+1/10) h = 16/60 h = 4/15 h`

If train 2 is to arrive at the same time as train 1, it must make the trip with an average speed of

`v_"ave" = (13 km)/(4/15 h) = 48.75 (km)/h`

Let `a_a, a_d, t_a, and t_d` be the accelerating and decelerating values of the acceleration and times, respectively. Also, let `v_"max"` be the max velocity we are looking for.

Using the kinematic formula `v = u + a*t`

During the accelerating and decelerating phases, we have these 2 formulas

`v_"max" = 0 + a_a*t_a`

`0 = v_"max" + a_d*t_d`

Solving both for `v_"max"`, and setting the 2 expressions on the other side of the equal sign equal to each other

`a_a*t_a = -a_d*t_d` `rarr` Eq 1

Let's get both accelerations on the left side and both times on the right.

`(a_a*cancel(t_a))/(a_d*cancel(t_a)) = -(cancel(a_d)*t_d)/(cancel(a_d)*t_a)`

So we have `(a_a)/(-a_d) = (t_d)/(t_a)`

It looks like if we double `a_a`, trying to get a higher `v_"max"`, we have to double `a_d` in order to stop at the other station. But wait! We also have `v_"ave" = 48.75 (km)/h`. Doubling both accelerations would decrease the time of the trip. The total time must be `t_a + t_d = 4/15 h` because both trains are to stop at the same moment.

I now see a path to the answer. Much of the work above was necessary, and perhaps interesting, so I will leave it there. If the accelerating part of the trip has an average speed

`v_"ave" = 48.75 (km)/h`,

then since the acceleration is uniform,

`v_"max" = 2*48.75 (km)/h = 97.5 (km)/h`.

I multiplied `v_"ave"` by 2 and said that is `v_"max"`. I need to defend that statement because it is key to the answer I foresee.

Picture a velocity time chart for the acceleration phase (or the deceleration phase). It is a right triangle. `v_"ave"` is halfway to the peak and it comes halfway thru the time. The area of this, in fact any, triangle is `("base"*"height")/2`. This is in units of velocity*time ... which is distance. It is in fact distance traveled with velocity and time being as plotted.

Now consider doing the entire trip at a speed of `v_"ave"`. That would be a rectangle with same base as the triangle but 1/2 the height. Same area as the one previously discussed. Therefore, play any games you like with greater acceleration. You would have to switch to deceleration when reaching a peak velocity of `97.5 (km)/h`. And then decelerate to a stop in the correct time. You would have 2 right triangles with the peak at `97.5 (km)/h`.

So the answer is `97.5 (km)/h`.

I hope this helps,
Steve