Answer

Verified

79.2k+ views

**Hint:**Above problem is related to the motion of the body in a rectilinear way. So, we can easily apply the equations of motion to solve this problem.

The motion of the body in which the body travels along a straight line is called rectilinear motion. This motion is an example of translatory motion.

**Formula used:**To solve rectilinear problems, we use equations of motion.

There are three equations of motion by which we can find initial velocity$\left( u \right)$, final the velocity$\left( v \right)$, acceleration$\left( a \right)$, displacement$\left( s \right)$ and time taken$\left( t \right)$ by the bodies in their travelling path. The equations of motion are following-

i)$v = u + at$

ii) $s = ut + \dfrac{1}{2}a{t^2}$

iii) ${v^2} = {u^2} + 2as$

**Complete step by step solution:**

In the given question, we have a body, which travels \[200cm\] in first \[2sec\]. So, for finding initial velocity, we can apply the second equation of motion because we know the displacement and time. By the second equation of motion-

$s = ut + \dfrac{1}{2}a{t^2}$

Substituting $s = 200cm$, $t = 2\sec $, we get-

$

200 = u \times 2 + \dfrac{1}{2}a{(2)^2} \\

\Rightarrow 200 = 2u + 2a \\

$

Or we can write-

$u + a = 100$ (i)

Now, in the next $4\sec $ body displaced $220cm$ more. So, the total displacement of the body will be $420cm(200cm + 220cm)$ and the total time taken $6\sec (2\sec + 4\sec )$. So, again by second equation of motion-

$s = ut + \dfrac{1}{2}a{t^2}$

Substituting $s = 420cm$, $t = 6\sec $, we get-

$

420 = u \times 6 + \dfrac{1}{2}a{(6)^2} \\

\Rightarrow 420 = 6u + 18a \\

$

Or we can write-

$u + 3a = 70$ (ii)

On simplifying equation (i) and equation (ii), we get-

$a = - 15cm/{\sec ^2}$

Putting the value in $a$ in equation (i)

$

u - 15 = 100 \\

\Rightarrow u = 100 + 15 \\

\Rightarrow u = 115cm/s \\

$

In the new case, question is asking for velocity at the end of the \[{7^{th}}sec\]-

Now, using first equation of motion-

$v = u + at$

Substituting $u = 115cm/s$, $a = - 15cm/{\sec ^2}$and $t = 7\sec $, we get-

$

v = 115 + \left( { - 15} \right) \times 7 \\

\Rightarrow v = 115 - 105 \\

\therefore v = 10cm/s \\

$

Hence, the velocity of the body at the end of \[{7^{th}}sec\] is $10cm/s$.

**Therefore, option (C) is correct.**

**Note:**We have to remember all equations of motion to solve this question. Here, we use the second equation of motion. If there is a change in the acceleration then we can use the third equation of motion. One thing is to be remembered then when the velocity is increasing then there is acceleration and if velocity is decreasing then there is deceleration or retardation.

Recently Updated Pages

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main

What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main

A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main

Other Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

The IUPAC name of the following compound is A Propane123tricarbonitrile class 12 chemistry JEE_Main

Choose the correct statements A A dimensionally correct class 11 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main