Question: Two trams, each of length 100 m are travelling in opposite direction with speed \( 15\;{\text{m}}/{\...

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

Solution

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.