
The initial velocity of a particle is $10 \mathrm{~m} / \mathrm{sec}$ and its retardation is $2 \mathrm{~m} / \mathrm{sec}^{2}$. The distance covered in the fifth second of the motion will be
(1) $1 \mathrm{~m}$
(2) $19 \mathrm{~m}$
(3) $50 \mathrm{~m}$
(4) $75 \mathrm{~m}$
Answer
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Hint: Equations of motion are equations that define a physical system 's action as a function of time in terms of its motion. More precisely, in terms of dynamic variables, the equations of motion describe the behavior of a physical system as a series of mathematical functions. To calculate the distance traveled during the nth second, you calculate the distance covered in n seconds and subtract the distance covered formula.
Formula used:
$\mathrm{S}_{\mathrm{n}}=\mathrm{u}+\dfrac{\mathrm{a}}{2}(2 \mathrm{n}-1)$
$\mathrm{S}_{\mathrm{n}}=\mathrm{u}+\dfrac{\mathrm{a}}{2}(2 \mathrm{n}-1)$= Distance travel in n second
$\mathrm{S}_{\mathrm{n}}=\mathrm{u}+\dfrac{\mathrm{a}}{2}(2 \mathrm{n}-1)$= Initial velocity of the body
$\text{a}$= acceleration of the body
Complete answer:
Given in the question:
The initial velocity of the particle $\mathrm{u}=10 \mathrm{~m} / \mathrm{~s}$
The retardation of the particle $\mathrm{a}=-2 \mathrm{~m} / \mathrm{~s}^{2}$
Time taken by the particle $\mathrm{n}=5$
The distance covered by the particle in $\mathrm{n}^{\text {th }}$ second is
$\mathrm{S}_{\mathrm{n}}=\mathrm{u}+\dfrac{\mathrm{a}}{2}(2 \mathrm{n}-1)$
$\Rightarrow {{\text{S}}_{5}}=10-\dfrac{2}{2}(2\times 5-1)$
$\Rightarrow {{\text{S}}_{5}}=10-9$
$\therefore {{\text{S}}_{5}}=1~\text{m}$
\[\therefore \] The distance covered in the fifth second of the motion will be $1~\text{m}$
Option (1) is correct.
Note:
The motion equations of kinematics define the most basic principles of an object's motion. In 1D, 2D and 3D, these equations govern the motion of an object. They can easily be used at different times to measure expressions such as an object's position, velocity, or acceleration. The second motion equation gives the displacement of an object under continuous acceleration. There is also a corollary in the second motion equation, which gives the distance travelled by a body in n seconds of motion. An object's motion can follow many different paths. Here in a straight line (one dimension), we will focus on motion.
Formula used:
$\mathrm{S}_{\mathrm{n}}=\mathrm{u}+\dfrac{\mathrm{a}}{2}(2 \mathrm{n}-1)$
$\mathrm{S}_{\mathrm{n}}=\mathrm{u}+\dfrac{\mathrm{a}}{2}(2 \mathrm{n}-1)$= Distance travel in n second
$\mathrm{S}_{\mathrm{n}}=\mathrm{u}+\dfrac{\mathrm{a}}{2}(2 \mathrm{n}-1)$= Initial velocity of the body
$\text{a}$= acceleration of the body
Complete answer:
Given in the question:
The initial velocity of the particle $\mathrm{u}=10 \mathrm{~m} / \mathrm{~s}$
The retardation of the particle $\mathrm{a}=-2 \mathrm{~m} / \mathrm{~s}^{2}$
Time taken by the particle $\mathrm{n}=5$
The distance covered by the particle in $\mathrm{n}^{\text {th }}$ second is
$\mathrm{S}_{\mathrm{n}}=\mathrm{u}+\dfrac{\mathrm{a}}{2}(2 \mathrm{n}-1)$
$\Rightarrow {{\text{S}}_{5}}=10-\dfrac{2}{2}(2\times 5-1)$
$\Rightarrow {{\text{S}}_{5}}=10-9$
$\therefore {{\text{S}}_{5}}=1~\text{m}$
\[\therefore \] The distance covered in the fifth second of the motion will be $1~\text{m}$
Option (1) is correct.
Note:
The motion equations of kinematics define the most basic principles of an object's motion. In 1D, 2D and 3D, these equations govern the motion of an object. They can easily be used at different times to measure expressions such as an object's position, velocity, or acceleration. The second motion equation gives the displacement of an object under continuous acceleration. There is also a corollary in the second motion equation, which gives the distance travelled by a body in n seconds of motion. An object's motion can follow many different paths. Here in a straight line (one dimension), we will focus on motion.
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