If the engine of a long train moving with con | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Using the third equation of motion,$v_f^2 - v_i^2 = 2as$. Let the total length of the train be $L$. When the engine crosses the tree,the initial v

Vedclass · Accepted Answer

$\sqrt{\frac{v^2+u^2}{2}}$

Vedclass · Answer

(C) Using the third equation of motion,$v_f^2 - v_i^2 = 2as$. Let the total length of the train be $L$. When the engine crosses the tree,the initial velocity is $u$. When the last compartment crosses the tree,the train has covered a distance equal to its length $L$,and its final velocity is $v$. Thus,$v^2 - u^2 = 2aL$,which gives $a = \frac{v^2 - u^2}{2L}$ ....$(1)$ Now,consider the middle compartment. It is at a distance of $\frac{L}{2}$ from the engine. Let $v_m$ be the velocity of the middle compartment as it crosses the tree. Using the equation of motion for the distance $\frac{L}{2}$: $v_m^2 - u^2 = 2a(\frac{L}{2}) = aL$. Substituting the value of $a$ from equation $(1)$: $v_m^2 = u^2 + (\frac{v^2 - u^2}{2L}) 	imes L$ $v_m^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{2u^2 + v^2 - u^2}{2} = \frac{v^2 + u^2}{2}$ Therefore,$v_m = \sqrt{\frac{v^2 + u^2}{2}}$.

If the engine of a long train moving with constant acceleration crosses a tree with velocity $u$ and the last compartment of the train crosses the same tree with velocity $v$,then the velocity with which the middle compartment crosses the same tree is

Similar Questions

The relation between time $t$ and displacement $x$ is $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. If $v$ is the velocity,the retardation is:

$A$ body travels $102.5 \ m$ in the $n^{\text{th}}$ second and $115.0 \ m$ in the $(n+2)^{\text{th}}$ second. The acceleration is: (in $m/s^2$)

$A$ body starts from rest. What is the ratio of the distance travelled by the body during the $4^{th}$ and $3^{rd}$ second?

Vedclass Test Series

Exam Paper Generator

Online Exam Module