Calculating Total Distance Covered by a Subway Train in 84 Seconds

[LICACS]

A subway train starts from rest at a station and accelerates at a rate of 1.60m/s2 for 14.0s. It runs at a constant speed for 70.0s and slows down at a rate of 3.50m/s2 until it stops at the next station. Find the total distance covered.

I have no idea what to do for this problem.

 

[hage567]

What kinematic equations do you know? You must have some idea if you are studying this stuff. Give more information on what it is you don't understand. Start by breaking up the problem into pieces. The accelerating part, the constant speed part and the decelerating part.

Give it a try.

 

[Moetasim]

I would approach this problem by using the principles of physics and mathematical equations to calculate the total distance covered by the subway train.

First, I would identify the known variables in the problem: acceleration (1.60m/s2 and 3.50m/s2), time (14.0s and 70.0s), and initial velocity (0m/s).

Next, I would use the equation d = v0t + 1/2at^2 to calculate the distance covered during the acceleration phase. Since the train starts from rest, the initial velocity (v0) is 0m/s, and the acceleration (a) is 1.60m/s2, the equation becomes d = 0 + 1/2(1.60)(14.0)^2 = 156.8m.

Then, I would use the equation v = v0 + at to calculate the final velocity after the acceleration phase. The initial velocity is still 0m/s, and the acceleration is 1.60m/s2, so the final velocity (v) becomes v = 0 + (1.60)(14.0) = 22.4m/s.

Using the final velocity, I can then calculate the distance covered during the constant speed phase using the equation d = vt. Since the train runs at a constant speed for 70.0s, the distance covered is d = (22.4)(70.0) = 1568m.

Finally, I can use the equation v^2 = v0^2 + 2ad to calculate the distance covered during the deceleration phase. The final velocity is 0m/s, the initial velocity is 22.4m/s, and the acceleration is -3.50m/s2 (negative because the train is slowing down). The equation becomes 0 = (22.4)^2 + 2(-3.50)d, and solving for d gives a distance of 319.36m.

Therefore, the total distance covered by the subway train in 84 seconds is 156.8m + 1568m + 319.36m = 2044.16m.