Question

From a station 'X' a train starts from rest and attains a speed of 54 km/h in 10s . Then by applying brakes negative acceleration of 2.5 m/s² is produced and it stops at station 'Y' in 6s . Find the distance between station 'X' and 'Y' ?​

Answer 1

Question :

From a station 'X' a train starts from rest and attains a speed of 54 km/h in 10s. Then by applying brakes negative acceleration of 2.5 m/s² is produced and it stops at station 'Y' in 6s . Find the distance between station 'X' and 'Y' ?​

Answer :

120 m

Explanation :

Part - 1 :

• From a station 'X' a train starts from rest and attains a speed of 54 km/h in 10s

          Initial speed, u = 0 m/s

          final speed, v = 54 km/h

                              v = 54 × (5/18) m/s

                              v = (3 × 5) m/s

                              v = 15 m/s

          time taken, t = 10 s

we know,

           $\boxed{\bf v=u+at}$

           15 = 0 + a(10)

            15 = 10a

            a = 15/10

            a = 1.5 m/s²

∴ acceleration = 1.5 m/s²

Distance travelled :

         $\boxed{\bf S=ut+\frac{1}{2}at^2}$

         $S_1=0(10) +\frac{1}{2}(1.5)(10^2) \\\\ S_1=\frac{1}{2} \times \frac{15}{10} \times 100 \\\\S_1=15 \times 5 \\\\ S_1=75 \ m$

Distance travelled in the first part = 75 m

Part - 2 :

• Then by applying brakes negative acceleration of 2.5 m/s² is produced and it stops at station 'Y' in 6s

                 for this part,

           initial velocity will be 54 km/h i.e., v = 15 m/s

            retardation, a = 2.5 m/s²

            time taken, t = 6 s

Distance travelled :

                $\boxed{\bf S=vt-\frac{1}{2}at^2}$                [retardation]

                 $S_2=15(6)-\frac{1}{2}(2.5)(6^2) \\\\S_2=90-\frac{1}{2} (\frac{25}{10})(36) \\\\ S_2=90-(5 \times 9) \\\\ S_2=90-45\\\\S_2=45 \ m$

Distance travelled in the second part = 45 m

⇒ The distance between station 'X' and 'Y'

        $=S_1+S_2 \\\\ =75+45\\\\=120 \ m$

Answer 2

$\bold{To\:find\::}$ Distance between station X & Y

$\bold{Given\::}$

$\bullet$ At station X , train was at rest

$\bullet$ velocity of train 10s after start of journey = 54km/h

$\bullet$ negative acceleration, after applying brake = 2.5m/s²

$\bullet$ time for which negative acceleration is applied before reaching Y = 6s

$\bold{Let\::}$

$\bullet$ distance covered by train in first 10s = A

$\bullet$ distance covered in next 6s = B

$\bold{Formula\:used\::}$

$\bullet$ v = u + a×t

$\bullet$ S = u×t + (1/2)×a×t²

$\bullet$ v² - u² = 2×a×S

$\bullet$ 1km/h = $\dfrac{5}{18}$m/s

$\bullet$ distance between station X &Y = A+B

$\bold{Solution\::}$

$\bullet$ For first $\bold{10s\::}$

$\sf u_{1}$ = 0

$\sf v_{1}$ = 54km/hr = 54× $\dfrac{5}{18}$m/s = 15m/s

$\sf t_{1}$ = 10s

$\sf S_{1}$ = A

Using identity, v = u + a×t

$\implies$ $\sf v_{1}$ = $\sf u_{1}$ + $\sf a_{1}$ × $\sf t_{1}$

putting respective values,

$\implies$ 15m/s = 0 +  $\sf a_{1}$× 10s

$\implies$ $\sf a_{1}$ = 15 ÷ 10 = 1.5m/s²

using identity, S = u×t + (1/2)×a×t²

Putting respective values

$\implies$ A = (0)×(10) + (1/2)×(1.5)×(10²)

$\implies$ A = (1/2)×(1.5)×(100) = (1/2)×(150)

$\implies$ A = 75m

$\bullet$ For next $\bold{6s\::}$

$\sf u_{2}$ =  54km/hr = 54× $\dfrac{5}{18}$m/s = 15m/s

$\sf v_{2}$ = 0

$\sf t_{2}$ = 6s

$\sf S_{2}$ = B

$\sf a_{2}$ = -2.5m/s² ( as it is negative retardation)

Using identity, v² - u² = 2×a×S

Putting respective values,

$\implies$ 0² - (15)² = 2 × (-2.5) × B

$\implies$ (- 225) = (-5) × B

$\implies$ B = (-225) ÷ (-5)

$\implies$ B = 45m

Therefore, distance between station X & Y = A+B

Putting respective values,

Distance between station X &Y = 75m + 45m

Distance between station X & Y = $\bold{120m\::}$

$\bold{ANSWER\::}$ $\bold{120m\::}$