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?

1 Answer
May 23, 2018

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