Question

two trains each of length 100m are travelling in opposite direction with speed 15m/s and 25m/s.The time taken for crossing?

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Time\:taken=</p><p>2.5\:sec}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about two trains each of length 100m are travelling in opposite direction with speed 15m/s and 25m/s.

• We have to find the time taken for crossing.

$\underline \bold{Given : } \\ \implies Distance(s) = 100 \: m \\ \\ \implies Velocity \: of \: train \: A = 15 \: m/s \\ \\ \implies Velocity \: of \: train \: B = 25\: m/s \\ \\\underline \bold{To \: Find : } \\ \implies Time \: taken \: to \: cross = ?$

• According to given question :

$\bold{Solving \: by \: Relative \: motion : } \\ \\ \bold{Assume \:train \: A \: in \: rest} \\ \implies Initial \: velocity = 0 \\ \\ \bold{According \: to \: Relative \: motion} \\ \implies Velocity \: of \: train \: B= 25+ 15 \\ \\ \bold{\implies Velocity \: of \: train \: B = 40\: m/s} \\ \\ \bold{For \: finding \: time : } \\ \implies Time = \frac{Distance}{Speed} \\ \\ \implies Time = \frac{ \cancel{100}}{ \cancel{40}} \\ \\ \bold{\implies Time = 2.5 \: sec} \\ \\ \bold{Alternate\: method : } \\ \\ \bold{For \: train \: A} \\ \implies s = ut + \frac{1}{2} a {t}^{2} \\ \\ \implies x = 15 \times t + \frac{1}{2} \times 0 \times {t}^{2} \\ \\ \implies x = 15t - - - - - (1) \\ \\ \bold{For \: train \: B} \\ \implies s = ut + \frac{1}{2} a {t}^{2} \\ \\ \implies 100 - x = 25 \times t + 0 - - - - - (2)\\ \\ \bold{Putting \: value \: of \: x \: in \: (2)} \\ \implies 100 - 15t = 25t \\ \\ \implies 100 = 40t \\ \\ \implies t = \frac{ \cancel{100}}{ \cancel{40}} \\ \\ \bold{\implies t = 2.5 \: sec}$

Answer 2

Answer:

○ Time taken=2.5 sec

Step-by-step explanation :

given:

□ Distance=100 m

□ Velocity of train A=15 m/s

□ Velocity of train B= 25 m/s

For train A--

Let distance travelled x m.

Using third eqn of of motion:

$\to s = ut + \frac{1}{2} a {t}^{2} \\ \\ \to x = 15t + \frac{1}{2} {at}^{2} - - - - - (1)$

For train B--

Using third eqn of of motion:

$\to s = ut + \frac{1}{2} {at}^{2} \\ \\ \to 100 - x = 25t + \frac{1}{2} {at}^{2} - - - - - (2) \\ \\ \bold{substituting \: value \: of \: x \: in \: eqn \: (2)} \\ \\ \to 100 - 15t + \frac{1}{2} {at}^{2} = 25t + \frac{1}{2} {at}^{2} \\ \\ \to 100 = 25t + 15t \\ \\ \to 40t = 100 \\ \\ \to t = \frac{100}{40} \\ \\ \to t = 2.5 \: sec$