A train is travelling at 10 m s$^{-1}$ on a straight horizontal track - Edexcel - A-Level Maths Mechanics - Question 5 - 2005 - Paper 1

To calculate the distance travelled during the deceleration:

1. Use the formula for distance under constant acceleration: $d = vt + \frac{1}{2}at^2$
2. First, determine the deceleration:
   • Initial speed (u) = 10 m s$^{-1}$
   • Final speed (v) = 3 m s$^{-1}$
   • Time (t) = 12 s
   • Using the equation: $v = u + at$ we can rearrange to find a: $a = \frac{v - u}{t} = \frac{3 - 10}{12} = -\frac{7}{12} m s^{-2}$
3. Now substitute into the distance formula:
   • Distance: $d = (10)(12) + \frac{1}{2}(-\frac{7}{12})(12^2)$ $d = 120 - 42 = 78 m$

1. After the initial braking, the train moves at a constant speed of 3 m s$^{-1}$ for 15 seconds.

2. Distance remaining to stop after 12 seconds:

   • Total distance to signal = 135 m
   • Distance travelled during initial deceleration = 78 m
   • Distance remaining = 135 m - 78 m = 57 m

3. Now find the time taken to cover this remaining distance while decelerating from 3 m s$^{-1}$ to rest:

   • Using the equation: $d = vt + \frac{1}{2}at^2$

   • Since the train stops, final speed (v) = 0. The initial speed (u) = 3 m s$^{-1}$ and distance (d) = 57 m.

4. From $v = u + at$, we have: $0 = 3 + a t$ So, $a = -\frac{3}{t}$ and substituting for $d$ gives: $57 = 3t + \frac{1}{2}(-\frac{3}{t})t^2$

5. Solving this gives: $6t^2 + 57t - 6(57) = 0$

   • Use the quadratic formula to find t, which evaluates to t = 8 s.

6. Hence, total time = 27 s (first braking) + 8 s (final braking) = 35 s.