Answer:
Time that they collide = 4.99s
Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s
Explanation:
We will use the equations of motion to obtain the solution required
At time t = 0
speed of first train = 22.2 m/s
Initial space between the two trains = 50 m
Speed of second train = 6.94 m/s
For the first car, distance covered by the first train = y
y = distance covered between the beginning of the deceleration and the point where the the two trains hit one another.
u = initial velocity = 22.2 m/s
t = time taken for all this to happen
a = deceleration = - 2.1 m/s²
y = ut + (1/2)at²
y = 22.2t - 1.05t² (eqn 1)
For the second train,
At t = 0, y = 50 m
Let the new distance moved by the second train before collision = (y - 50)
u = initial velocity = 6.94 m/s
t = time taken = t
a = acceleration of the second train = 0 m/s² (constant velocity)
(y - 50) = ut + (1/2)at²
y - 50 = 6.94t
y = 6.94t + 50 (eqn 2)
substituting for y in eqn 2 using the expression obtained in eqn 1
y = 22.2t - 1.05t²
y = 6.94t + 50
22.2t - 1.05t² = 6.94t + 50
1.05t² - 15.26t + 50 = 0
Solving this quadratic equation
t = 4.99 s or 9.54 s
The position of the two trains are the same at those two times, but the first time is when they hit each other.
t = 4.99 s
At 4.99 s, the the velocity of the first train
v = u + at
v = 22.2 + (-2.1×4.99) = 11.721 m/s in the same direction as the second train.
Relative velocity at this point will be
= 11.721 - 6.94 = 4.781 m/s
Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s
Hope this Helps!!!
