Question

Two trams, each of length 100 m are travelling in opposite direction with speed $ 15\;{\text{m}}/{\text{s}} $ and $ 25\;{\text{m}}/{\text{s}} $ . The time taken for crossing is
(A) $ 4s $
(B) $ 2.5s $
(C) $ 5s $
(D) $ 2s $

Answer 1

Hint: We can easily solve this question by using the third equation of motion. The third equation of motion gives the relation between distance, speed, acceleration, and time.

Formula Used: We will use the following formula to find the answer to the question
 $ s = ut + \dfrac{1}{2}a{t^2} $
Where, $ s $ is the distance travelled,
 $ u $ is the initial velocity,
 $ a $ is the acceleration of the body,
 $ t $ is the time taken by the body.

Complete Step-by-Step Solution
According to the question, the following information is provided to us
The length of each tram is $ 100m $
Velocity of tram $ A = 15m/s $
Velocity of tram $ B = 25m/s $
Now, let us suppose that the distance travelled by the tram $ A $ be $ x $ metres.
Then, the distance travelled by tram $ B $ will be $ (100 - x) $ metres.
Let us now use the third equation of motion for tram $ A $ .
 $ s = ut + \dfrac{1}{2}a{t^2} $
Now we will replace $ s $ with $ x $ to get
 $ x = 15t + \dfrac{1}{2}a{t^2} $
Similarly, we will use the third equation of motion for tram $ B $ .
 $ s = ut + \dfrac{1}{2}a{t^2} $
Now we will substitute the values of $ s $ and $ u $ in the above equation to get
 $ 100 - x = 25 \times t + \dfrac{1}{2}a{t^2} $
Now, we will substitute the value of $ x $ that we found out from the earlier equation to this equation.
That is,
 $ 100 - 15t + \dfrac{1}{2}a{t^2} = 25t + \dfrac{1}{2}a{t^2} $
 $ \Rightarrow 100 = 25t + 15t $
Upon further solving, we get
 $ \Rightarrow 40t = 100 $
 $ \Rightarrow t = \dfrac{{100}}{{40}} $
Therefore, $ t = 2.5\sec $
Hence, the correct option is (B).

Note
We can also solve this question with the concept of relative motion. That method will be a little bit difficult to understand but it is a very short method. Motion as seen from or referred to a certain material system that constitutes a reference frame (as two adjacent walls and floor of a room) is known as Relative Motion.