
Particle A starts with initial velocity 10 m/s and acceleration $1m/{s^2}$. From the same point another particle B starts with initial velocity 5 m/s and acceleration $2{\text{ }}m/{s^2}$. Both the particles reach the destination at the same time. Find the length of the path
A) 200m
B) 150m
C) 100m
D) 50m
Answer
554.7k+ views
Hint: The second equation of motion also known as the position time relation is used to calculate the distance travelled by the body. Find the expression of the distance travelled by both the particles then equate them to find the required answer.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
Complete answer:
The second equation of motion also known as the position time relation is mathematically given as, $S = ut + \dfrac{1}{2}a{t^2}$
Where ‘s’ is the displacement, ‘u’ is the initial velocity, ‘t’ is the time and ‘a’ is the acceleration of the body.This formula is used to calculate the distance when the values for time, acceleration and initial velocity are known.
Write down all the given physical quantities, then apply the formula of the second equation of motion for each particle.
Particle A:
${u_1} = 10$, ${a_1} = 1$
Particle B:
${u_2} = 5$, ${a_2} = 2$
Distance travelled by particle A is given by,
${s_1} = {u_1}t + \dfrac{1}{2}{a_1}{t^2}$
$ \Rightarrow {s_1} = 10t + \dfrac{1}{2}{t^2}$ --(1)
Distance travelled by particle B is given by,
${s_2} = {u_2}t + \dfrac{1}{2}{a_2}{t^2}$
$ \Rightarrow {s_1} = 5t + {t^2}$
According to the question the particle A and particle B start from the same point and reach the destination at the same time, which means that the particles are travelling the same path and since both of them reached the destination at the same time that means the time taken by both the particles is equal.
i.e. ${s_1} = {s_2}$
$ \Rightarrow 10t + \dfrac{1}{2}{t^2} = 5t + {t^2}$
$ \Rightarrow 5t = \dfrac{{{t^2}}}{2}$
$ \Rightarrow t = 10$
Substitute the value of time $t = 10$ in equation (1) we get,
$ \Rightarrow {s_1} = 10t + \dfrac{1}{2}{t^2}$
$ \Rightarrow {s_1} = 10\left( {10} \right) + \dfrac{1}{2}{\left( {10} \right)^2}$
$ \Rightarrow {s_1} = 100 + 50 = 150m$
Since, the path travelled by both particle A and B is equal,
Thus, ${s_1} = {s_2}$ = 150m
Therefore, the length of the path travelled by particle A and particle B is 150m.
So, the correct answer is “Option B”.
Note:
Equations of motion are the mathematical formulas which describe the position, velocity, or acceleration of a body relative to a given frame of reference. In physics, motion is defined as the change in the position of the body with respect to time. Motion which is along a straight line or curve is known as translational motion and motion which changes the orientation of the body is known as the rotational motion.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
Complete answer:
The second equation of motion also known as the position time relation is mathematically given as, $S = ut + \dfrac{1}{2}a{t^2}$
Where ‘s’ is the displacement, ‘u’ is the initial velocity, ‘t’ is the time and ‘a’ is the acceleration of the body.This formula is used to calculate the distance when the values for time, acceleration and initial velocity are known.
Write down all the given physical quantities, then apply the formula of the second equation of motion for each particle.
Particle A:
${u_1} = 10$, ${a_1} = 1$
Particle B:
${u_2} = 5$, ${a_2} = 2$
Distance travelled by particle A is given by,
${s_1} = {u_1}t + \dfrac{1}{2}{a_1}{t^2}$
$ \Rightarrow {s_1} = 10t + \dfrac{1}{2}{t^2}$ --(1)
Distance travelled by particle B is given by,
${s_2} = {u_2}t + \dfrac{1}{2}{a_2}{t^2}$
$ \Rightarrow {s_1} = 5t + {t^2}$
According to the question the particle A and particle B start from the same point and reach the destination at the same time, which means that the particles are travelling the same path and since both of them reached the destination at the same time that means the time taken by both the particles is equal.
i.e. ${s_1} = {s_2}$
$ \Rightarrow 10t + \dfrac{1}{2}{t^2} = 5t + {t^2}$
$ \Rightarrow 5t = \dfrac{{{t^2}}}{2}$
$ \Rightarrow t = 10$
Substitute the value of time $t = 10$ in equation (1) we get,
$ \Rightarrow {s_1} = 10t + \dfrac{1}{2}{t^2}$
$ \Rightarrow {s_1} = 10\left( {10} \right) + \dfrac{1}{2}{\left( {10} \right)^2}$
$ \Rightarrow {s_1} = 100 + 50 = 150m$
Since, the path travelled by both particle A and B is equal,
Thus, ${s_1} = {s_2}$ = 150m
Therefore, the length of the path travelled by particle A and particle B is 150m.
So, the correct answer is “Option B”.
Note:
Equations of motion are the mathematical formulas which describe the position, velocity, or acceleration of a body relative to a given frame of reference. In physics, motion is defined as the change in the position of the body with respect to time. Motion which is along a straight line or curve is known as translational motion and motion which changes the orientation of the body is known as the rotational motion.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Trending doubts
In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

