Question

A body covers $200\,cm$ in the first 2 second and $220\,cm$ in the next 4 seconds. What is the velocity of the body at the end of ${7^{th}}$ second.
A. $40\,cm/s$
B. $20\,cm/s$
C. $10\,cm/s$
D. $5\,cm/s$

Answer 1

Hint-
For solving this question, we can use the equation of motion, $v = u + at$ to find the final velocity v. But the value of initial velocity and acceleration is not given. For this we can use the second equation of motion
$S = ut + \dfrac{1}{2}a{t^2}$
Where S denotes the distance travelled, u denotes the initial velocity t is the time taken and a is the acceleration.
By writing the equation of distance for the first 2 seconds and first 6 seconds and solving we can find the initial velocity and acceleration.

Step by step solution:
Let the distance travelled in 2 seconds be ${S_1}$ and the distance travelled in the next 4 seconds be ${S_2}$
It is given that the initial distance is $200\,cm$. Therefore,
${S_1} = 200\,cm$
Time ${t_1} = 2\,s$
Distance travelled in next 4 second is given as $220\,cm$ . therefore,
${S_2} = 220\,cm$
${t_2} = 4\,s$
Now we need to find the velocity at the end of ${7^{th}}$ second.
For this let us use the equation of motion.
We know that according to the equation of motion.
$S = ut + \dfrac{1}{2}a{t^2}$
Where S denotes the distance travelled, u denotes the initial velocity t is the time taken and a is the acceleration.
In the first case we can write,
${S_1} = u{t_1} + \dfrac{1}{2}a{t_1}^2$
On substituting the values we get
$200 = u \times 2 + \dfrac{1}{2}a \times 4$
$\Rightarrow 200 = 2u + 2a$
$\Rightarrow 100 = u + a$....................(1)
Now let us write the equation for the total distance travelled in 6 seconds..
c
Time $t = 6\,s$
$S = {S_1} + {S_2} = 200 + 220 = 420\,cm$
On substituting the values given we get ,
$420 = u \times 6 + \dfrac{1}{2}a \times 36$
$\Rightarrow 420 = 6u + 18a$
$\Rightarrow 70 = u + 3a$ …………………...(2)
Now let us solve equation (1) and (2) .
On subtracting equation (2) from equation (1), We get
$30 = - 2a$
$\therefore a = - 15\,cm/{s^2}$
On adding equation 1 and 2 .
$170 = 2u + 4a$
On substitute the value of a we get
$170 = 2u - 60$
$\Rightarrow u = 115\,cm/s$
Using this we can find the velocity at seventh second.
Let us use the first equation of motion.
$v = u + at$
Substitute the value of time as seven . then we get
$v = 115 + - 15 \times 7$
$v = 10\,cm/s$
This is the velocity at seventh second.

So the answer is option C.

Note: Remember to substitute the total distance travelled in the first two second and next four second to get the total distance covered in the first 6 seconds. In the question the distance of the first 2 seconds and next 4 second is given. In order to get the value of initial velocity and acceleration we wrote the equation of motion involving distance for the first two seconds and for the first 6 seconds. Then only we can take the initial velocity to be the same in both equations so that we can solve those equations to get the value of initial velocity.