Some how TRAIN A travelling at 36 m/s, is accidentally sidetracked onto the train track for TRAIN A. The TRAIN A engineer spots TRAIN A 100 m ahead on the same track and travelling in the same direction. The engineer jams on the brakes and slows TRAIN B at a rate of 3m/s^2. TRAIN A's conductor was unaware of the situation. If the speed of the local train is 11 m/s will the express be able to stop in time or will there be a collision?

Support you answer with a calculation.

Please help me solve this ASAP, thank you.

Sorry Train A is sidetracked onto Train B's train track, which is 100 m ahead of Train A

the answer is 12 seconds. you divide 36 by 3 not 2.

To solve this problem, we need to analyze the motion of both trains and determine if there is enough distance for the express train (TRAIN B) to stop before colliding with the local train (TRAIN A).

First, let's define the variables:
- Initial velocity of TRAIN B (express train) = 36 m/s
- Distance between TRAIN B and TRAIN A = 100 m
- Initial velocity of TRAIN A (local train) = 11 m/s
- Acceleration of TRAIN B (braking rate) = -3 m/s² (negative because it represents deceleration)

We can use the following equation of motion for TRAIN B to determine whether it will collide with TRAIN A:

v^2 = u^2 + 2as

where:
- v is the final velocity (0 m/s, since TRAIN B needs to stop)
- u is the initial velocity of TRAIN B (36 m/s)
- a is the acceleration (braking rate) of TRAIN B (-3 m/s²)
- s is the distance between TRAIN B and TRAIN A (100 m)

Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Substituting the values into the equation:

s = (0^2 - 36^2) / (2 * -3)
s = (-1296) / (-6)
s = 216 m

Therefore, the distance required for TRAIN B to stop is 216 m. However, the distance between TRAIN B and TRAIN A is only 100 m. Since the stopping distance is greater than the available distance, TRAIN B will not be able to stop in time and there will be a collision.

Note: It's essential to keep in mind that this calculation assumes constant acceleration and neglects other factors like reaction time and the time it takes for the brakes to engage. In a realistic scenario, additional considerations might be necessary.

v=0

u=36m/s
a=3m/s^2
t=?
v=u+at
0=36m/s-3m/s^2t
0=-36=3t
t=36/2
=18 seconds.