
A particle starts from rest and moves with uniform acceleration. If covers a displacement of ${y^2} - {x^2}$ in the first $10s$ and ${y^2} + {x^2}$ in the next first $10s$, then:
A) $x = \sqrt 2 $
B) $x = 3y$
C) $y = 3x$
D) $y = \sqrt 2 x$
Answer
232.8k+ views
Hint: In the question, the displacement of the particle in the next first $10s$ implies that the time taken for the displacement is $20s$. Now, applying the equations of motion, two relations between displacement and time can be obtained. Solving them we can obtain the required solution.
Complete step by step solution:
From the question, we know that the particle moves with uniform acceleration. This means it travels an equal distance in equal intervals of time.
Since, the acceleration of the body remains constant throughout the motion, we can say the acceleration of the body in the first $10s$ and in the next first $10s$ are the same.
Now, using the third equation of motion, we know:
$S = ut + \dfrac{1}{2}a{t^2}$
Where:
$S$ is the displacement of the body
$u$ is the initial velocity of the body
$a$ is the acceleration of the body
$t$ is the time taken by the body to cause the given displacement.
As in the question, it is given that the body starts from rest, we can write that the initial velocity is zero.
Therefore, $u = 0$
Therefore, the equation of motion modifies to:
$\Rightarrow S = \dfrac{1}{2}a{t^2}$
Now, it is given, that displacement is ${y^2} - {x^2}$ in the first $10s$ , so we write:
$\Rightarrow {y^2} - {x^2} = \dfrac{1}{2}a{(10)^2}$
$ \Rightarrow {y^2} - {x^2} = 50a.......(1)$
And the displacement is ${y^2} + {x^2}$in the next first$10s$, therefore, similarly we write:
$\Rightarrow {y^2} + {x^2} + {y^2} - {x^2} = \dfrac{1}{2}a{(20)^2}$
$ \Rightarrow 2{y^2} = 200a$
$ \Rightarrow {y^2} = 100a......(2)$
Now, solving equation $(1)$ and$(2)$, we obtain:
$\Rightarrow {y^2} = 2{x^2}$
Therefore, we can write:
$\Rightarrow y = \sqrt 2 x$
This is the required solution.
Hence, option (D) is correct.
Note: The equation of motions defines the behavioural motion of a particle or body with respect to time. Using these equations of motion, we can draw the graphs of the motion. This enables us to obtain various quantities like acceleration, velocity, their instantaneous values and various other unknown quantities.
Complete step by step solution:
From the question, we know that the particle moves with uniform acceleration. This means it travels an equal distance in equal intervals of time.
Since, the acceleration of the body remains constant throughout the motion, we can say the acceleration of the body in the first $10s$ and in the next first $10s$ are the same.
Now, using the third equation of motion, we know:
$S = ut + \dfrac{1}{2}a{t^2}$
Where:
$S$ is the displacement of the body
$u$ is the initial velocity of the body
$a$ is the acceleration of the body
$t$ is the time taken by the body to cause the given displacement.
As in the question, it is given that the body starts from rest, we can write that the initial velocity is zero.
Therefore, $u = 0$
Therefore, the equation of motion modifies to:
$\Rightarrow S = \dfrac{1}{2}a{t^2}$
Now, it is given, that displacement is ${y^2} - {x^2}$ in the first $10s$ , so we write:
$\Rightarrow {y^2} - {x^2} = \dfrac{1}{2}a{(10)^2}$
$ \Rightarrow {y^2} - {x^2} = 50a.......(1)$
And the displacement is ${y^2} + {x^2}$in the next first$10s$, therefore, similarly we write:
$\Rightarrow {y^2} + {x^2} + {y^2} - {x^2} = \dfrac{1}{2}a{(20)^2}$
$ \Rightarrow 2{y^2} = 200a$
$ \Rightarrow {y^2} = 100a......(2)$
Now, solving equation $(1)$ and$(2)$, we obtain:
$\Rightarrow {y^2} = 2{x^2}$
Therefore, we can write:
$\Rightarrow y = \sqrt 2 x$
This is the required solution.
Hence, option (D) is correct.
Note: The equation of motions defines the behavioural motion of a particle or body with respect to time. Using these equations of motion, we can draw the graphs of the motion. This enables us to obtain various quantities like acceleration, velocity, their instantaneous values and various other unknown quantities.
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