
A body starts from rest and is uniformly accelerated for $30\sec $.The distance covered in first $10\sec $ is \[{x_1}\] , for next $10\sec $ is \[{x_2}\] and for last $10\sec $ is \[{x_3}\] the then the ratio of \[{x_1}:{x_2}:{x_3}\]?
Answer
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Hint: In this question we will apply the second equation of motion and with the help of it calculate the distance travelled in each of the $10\sec $ period. Then we will divide the three equations which will be obtained and this will give the required ratio. The second equation of motion is:$s = ut + \dfrac{1}{2}a{t^2}$.
Where, $s$ is the displacement, $u$ is the initial velocity, $t$ is the time taken and $a$ is the acceleration of the body.
Complete step by step answer:
According to the second equation of motion,
$s = ut + \dfrac{1}{2}a{t^2}$
The distance which is travelled by the body in the first $10\sec $ is,
${x_1} = 0 \times 10 + \dfrac{1}{2} \times a \times {10^2}$
${x_1} = \dfrac{1}{2} \times a \times 100$
${x_1} = 50a.......(1)$
Similarly, the distance which is travelled by the body in the next $10\sec $ is,
${x_2} = {s_{20}} - {s_{10}}$
${x_2} = \left( {0 \times 20 + \dfrac{1}{2} \times a \times {{20}^2}} \right) - 50a$
On further solving this, we get,
${x_2} = 200a - 50a$
${x_2} = 150a......(2)$
Similarly, the distance which is travelled by the body in the last $10\sec $ is,
${x_3} = {s_{30}} - {s_{20}}$
${x_2} = \left( {0 \times 30 + \dfrac{1}{2} \times a \times {{30}^2}} \right) - 200a$
On further solving this, we get,
${x_2} = 450a - 200a$
${x_2} = 250a.....(3)$
On comparing the three equations,
${x_1}:{x_2}:{x_3} = 50a:150a:250a$
On simplifying this,
${x_1}:{x_2}:{x_3} = 1:3:5$
So, the required ratio is ${x_1}:{x_2}:{x_3} = 1:3:5$.
Note: It is important to note that the three equations of motion are only valid for uniformly accelerated motion. These equations are not valid for a non-uniformly accelerated motion. This is because for a non-uniformly accelerated motion, we cannot associate with the inertial frame of reference. And we also know that, for a non-inertial frame of reference Newton's law is not valid. Hence, we can apply Newton's law of motion only when the body is uniformly accelerated.
Where, $s$ is the displacement, $u$ is the initial velocity, $t$ is the time taken and $a$ is the acceleration of the body.
Complete step by step answer:
According to the second equation of motion,
$s = ut + \dfrac{1}{2}a{t^2}$
The distance which is travelled by the body in the first $10\sec $ is,
${x_1} = 0 \times 10 + \dfrac{1}{2} \times a \times {10^2}$
${x_1} = \dfrac{1}{2} \times a \times 100$
${x_1} = 50a.......(1)$
Similarly, the distance which is travelled by the body in the next $10\sec $ is,
${x_2} = {s_{20}} - {s_{10}}$
${x_2} = \left( {0 \times 20 + \dfrac{1}{2} \times a \times {{20}^2}} \right) - 50a$
On further solving this, we get,
${x_2} = 200a - 50a$
${x_2} = 150a......(2)$
Similarly, the distance which is travelled by the body in the last $10\sec $ is,
${x_3} = {s_{30}} - {s_{20}}$
${x_2} = \left( {0 \times 30 + \dfrac{1}{2} \times a \times {{30}^2}} \right) - 200a$
On further solving this, we get,
${x_2} = 450a - 200a$
${x_2} = 250a.....(3)$
On comparing the three equations,
${x_1}:{x_2}:{x_3} = 50a:150a:250a$
On simplifying this,
${x_1}:{x_2}:{x_3} = 1:3:5$
So, the required ratio is ${x_1}:{x_2}:{x_3} = 1:3:5$.
Note: It is important to note that the three equations of motion are only valid for uniformly accelerated motion. These equations are not valid for a non-uniformly accelerated motion. This is because for a non-uniformly accelerated motion, we cannot associate with the inertial frame of reference. And we also know that, for a non-inertial frame of reference Newton's law is not valid. Hence, we can apply Newton's law of motion only when the body is uniformly accelerated.
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