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# A body travels $200cm$ in the first $2sec$ and $220cm$ in the next $4sec$ with deceleration. The velocity of the body at the end of the ${7^{th}}sec$ is: A) $20cm/s$B) $15cm/s$C) $10cm/s$D) $0cm/s$

Last updated date: 12th Sep 2024
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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.