Question

A body travels 200 cm in the first two seconds and 220 cm in the next four seconds. What will be the velocity at the end of the $7^{th}$ second from the start?
$\left( A \right)10cm/s$
$\left( B \right)20cm/s$
$\left( C \right)15cm/s$
$\left( D \right)5cm/s$

Answer 1

Apply the kinematic equation of motion, where you substitute the value of $u$ is the initial velocity, $t$ is the time, $g$ is the acceleration of gravity. Apply the kinematic equation to find the distance in the two seconds. Then find distance for six seconds. Then using the equations obtains the final velocity at seventh second.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
$u$ is the initial velocity, $t$ is the time, $a$ is the acceleration.

Complete step by step answer:
Motion equation helps to describe a body’s location, velocity or acceleration relative to frame of reference. The velocity can be derived from the newton equation by the method of integration. The study of motion of the bodies without considering the cause of motion is called kinematics. Displacement may or may not be equal to the path length travelled of an object. Distance to unit time is called speed. It is a scalar quantity. The body attains uniform motion along a straight line when that body is moving with uniform velocity. $u$ is the initial velocity, $t$ is the time, $a$ is the acceleration.
Let us apply the kinematic equation, $s = ut + \dfrac{1}{2}a{t^2}$
In the first 2 seconds, the distance given is $200m$ then
$\Rightarrow 200 = u \times 2 + \dfrac{1}{2} \times a \times {\left( 2 \right)^2}$
$\Rightarrow 100 = u + a - - - - - \left( 1 \right)$
Now for the next four seconds,
Then the total time will be $2 + 4 = 6\sec$
$\Rightarrow y = u \times 6 + \dfrac{1}{2}a \times {\left( 6 \right)^2}$
$\Rightarrow y = 6u + 18a - - - - - - \left( 2 \right)$
We know that, $220m$ was travelled between two and six seconds
$\Rightarrow y - 200 = 220$
$\Rightarrow y = 420$
Substitute this value in equation 2
$\Rightarrow 420 = 6u + 18a$
$\Rightarrow u + 3a = 70 - - - - - \left( 3 \right)$
Equating 1 and 2
$\Rightarrow a = - 15cm/{s^2}$
Now we can determine the value of initial velocity
$\Rightarrow u = 100 - \left( { - 15} \right) = 115cm/\sec$
Let us find the final velocity for seventh second
$\Rightarrow v = 115 + \left( { - 15} \right) \times 7 = 10cm/\sec$

Hence, the correct answer is option (A).

Note: The initial velocity will be zero if the motion starts from rest and the frame of reference should be the same. The velocity can be derived from the newton equation by the method of integration. A body is said to be in uniform motion along a straight line when that body is moving with uniform velocity.