Two trains A and B are approaching each other on a straight track, the former with a uniform velocity of 25 m/s and other with 15m/s, when they are 225 m a part brakes are simultaneously applied to both of them. The deceleration given by the brakes to thetrain B increases linearly with time by `0.3 m//s^(2)` every second, while the train A is given a uniform deceleration, (a) What must be the minimum deceleration of the train A so that the trains do not collide ? (b) What is the time taken by the trains to come to stop ?

Options

A. `5m//s^(2)`
B. `2.5m//s^(2)`
C. `1.5m//s^(2)`
D. `7.5m//s^(2)`

Answer

To solve the problem, we need to determine the minimum deceleration of train A so that the two trains do not collide, and then find the time taken by both trains to come to a stop.

### Step-by-Step Solution:

**Step 1: Understand the initial conditions.**
- Train A has an initial velocity \( u_A = 25 \, \text{m/s} \).
- Train B has an initial velocity \( u_B = 15 \, \text{m/s} \).
- The distance between the two trains when brakes are applied is \( d = 225 \, \text{m} \).

**Step 2: Determine the deceleration of train B.**
- The deceleration of train B increases linearly with time at a rate of \( 0.3 \, \text{m/s}^2 \).
- Therefore, the deceleration at time \( t \) is given by:
\[ a_B = -0.3t \, \text{m/s}^2 \]

**Step 3: Calculate the time taken for train B to stop.**
- The final velocity of train B when it stops is \( v_B = 0 \).
- Using the equation of motion:
\[ v_B = u_B + a_B \cdot t \]
Substituting the values:
\[ 0 = 15 - 0.3t^2 \]
Rearranging gives:
\[ 0.3t^2 = 15 \implies t^2 = \frac{15}{0.3} = 50 \implies t = \sqrt{50} \approx 7.07 \, \text{s} \]

**Step 4: Calculate the distance traveled by train B during this time.**
- Using the equation of motion:
\[ s_B = u_B \cdot t + \frac{1}{2} a_B \cdot t^2 \]
Substituting the values:
\[ s_B = 15 \cdot 7.07 + \frac{1}{2} (-0.3) \cdot (7.07)^2 \]
\[ s_B = 106.05 - 0.15 \cdot 50 \approx 106.05 - 7.5 = 98.55 \, \text{m} \]

**Step 5: Calculate the distance remaining for train A to cover.**
- The distance covered by train B is approximately \( 98.55 \, \text{m} \).
- Therefore, the distance remaining between the two trains when train B stops is:
\[ d' = 225 - 98.55 \approx 126.45 \, \text{m} \]

**Step 6: Determine the deceleration required for train A.**
- Train A must cover the distance \( d' \) in the same time \( t \) that train B takes to stop.
- Using the equation of motion for train A:
\[ s_A = u_A \cdot t + \frac{1}{2} a_A \cdot t^2 \]
We know \( s_A = 126.45 \, \text{m} \) and \( t = 7.07 \, \text{s} \):
\[ 126.45 = 25 \cdot 7.07 + \frac{1}{2} a_A \cdot (7.07)^2 \]
\[ 126.45 = 176.75 + \frac{1}{2} a_A \cdot 50 \]
Rearranging gives:
\[ \frac{1}{2} a_A \cdot 50 = 126.45 - 176.75 \implies \frac{1}{2} a_A \cdot 50 = -50.3 \]
\[ a_A = \frac{-50.3 \cdot 2}{50} = -2.012 \, \text{m/s}^2 \]

**Step 7: Calculate the time taken by train A to stop.**
- Using the equation of motion:
\[ v_A = u_A + a_A \cdot t \]
Setting \( v_A = 0 \):
\[ 0 = 25 - 2.012 \cdot t \]
\[ t = \frac{25}{2.012} \approx 12.43 \, \text{s} \]

### Final Answers:
(a) The minimum deceleration of train A must be approximately \( 2.012 \, \text{m/s}^2 \) to avoid a collision.

(b) The time taken by train A to come to a stop is approximately \( 12.43 \, \text{s} \).