Two bodies, initially separated by x m and having an initial velocity 2 m/s each, are moving towards each other along a straight line. If the rate of increase of their speeds is 3$m/{s^2}$ and 2$m/{s^2}$ respectively and the two meet after 4 seconds, find x.
Answer
634.5k+ views
Hint: In this question, we will first use the third equation of motion. by substituting the given values in the equation, will help us to get the required answer easily. Further we will discuss the basics of the equation of motion, for our better understanding.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
Complete answer:
As we can see, the given question is an example of the motion of a particle moving along a straight line with constant acceleration. So, here we can use the Kinematics equation of motion.
Now, let the two bodies be A and B with initial velocities ${u_1}$ and ${u_2}$ respectively, having values:
${u_1} = {u_2} = 2m/s$
Now, let the acceleration of the given two bodies A and B be ${a_1}$ and ${a_2}$ respectively, given as:
${a_1} = 3m/{s^2}$
${a_2} = 2m/{s^2}$
Let x be the initial separation between A and B and t be the time taken for A and B to meet.
Here, time is given as
$t = 4s$
Now, let ${x_1}$ and ${x_2}$ be the distance covered by the two bodies A and B respectively after time t such that
$x = {x_1} + {x_2}$
Further, using Kinematics equation of motion to find the values of ${x_1}$ and ${x_2}$
${x_1} = {u_1} \times t + \dfrac{1}{2}{a_1} \times {t^2}$
Now, we substitute the given values in the above equation, we get:
${x_1} = 2 \times 4 + \dfrac{1}{2} \times 3 \times {4^2}$
$ \Rightarrow {x_1} = 32m$
Now, we find the value of ${x_2}$
${x_2} = {u_2} \times t + \dfrac{1}{2}{a_2} \times {t^2}$
Now, we substitute the given values in the above equation, we get:
${x_1} = 2 \times 4 + \dfrac{1}{2} \times 2 \times {4^2}$
$ \Rightarrow {x_2} = 24m$
Now, we know that:
$x = {x_1} + {x_2}$
By substituting the observed values in the above equation, we get:
$\eqalign{& x = 32 + 24 \cr
& \therefore x = 56m \cr} $
Therefore, we get the required result i.e., we get the value of x.
Additional information:
As we know that the equations of motion are the equations which describe the behavior of any physical system in terms of its motion as a function of time. Further we can also say that the equations of motion tell us the behavior of any physical system as a set of mathematical functions in terms of different dynamic variables. Here, these dynamic variables are said to be generally spatial coordinates and time are used, but other variables are also possible, for example- momentum components and time.
Note:
Here we should remember that this question can be easily solved by knowing the equation of motion. We should also observe that the equation of motion is only applicable to the classical system not in the quantum system i.e., we cannot apply the equation of motion on waves or any quantum particle.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
Complete answer:
As we can see, the given question is an example of the motion of a particle moving along a straight line with constant acceleration. So, here we can use the Kinematics equation of motion.
Now, let the two bodies be A and B with initial velocities ${u_1}$ and ${u_2}$ respectively, having values:
${u_1} = {u_2} = 2m/s$
Now, let the acceleration of the given two bodies A and B be ${a_1}$ and ${a_2}$ respectively, given as:
${a_1} = 3m/{s^2}$
${a_2} = 2m/{s^2}$
Let x be the initial separation between A and B and t be the time taken for A and B to meet.
Here, time is given as
$t = 4s$
Now, let ${x_1}$ and ${x_2}$ be the distance covered by the two bodies A and B respectively after time t such that
$x = {x_1} + {x_2}$
Further, using Kinematics equation of motion to find the values of ${x_1}$ and ${x_2}$
${x_1} = {u_1} \times t + \dfrac{1}{2}{a_1} \times {t^2}$
Now, we substitute the given values in the above equation, we get:
${x_1} = 2 \times 4 + \dfrac{1}{2} \times 3 \times {4^2}$
$ \Rightarrow {x_1} = 32m$
Now, we find the value of ${x_2}$
${x_2} = {u_2} \times t + \dfrac{1}{2}{a_2} \times {t^2}$
Now, we substitute the given values in the above equation, we get:
${x_1} = 2 \times 4 + \dfrac{1}{2} \times 2 \times {4^2}$
$ \Rightarrow {x_2} = 24m$
Now, we know that:
$x = {x_1} + {x_2}$
By substituting the observed values in the above equation, we get:
$\eqalign{& x = 32 + 24 \cr
& \therefore x = 56m \cr} $
Therefore, we get the required result i.e., we get the value of x.
Additional information:
As we know that the equations of motion are the equations which describe the behavior of any physical system in terms of its motion as a function of time. Further we can also say that the equations of motion tell us the behavior of any physical system as a set of mathematical functions in terms of different dynamic variables. Here, these dynamic variables are said to be generally spatial coordinates and time are used, but other variables are also possible, for example- momentum components and time.
Note:
Here we should remember that this question can be easily solved by knowing the equation of motion. We should also observe that the equation of motion is only applicable to the classical system not in the quantum system i.e., we cannot apply the equation of motion on waves or any quantum particle.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

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

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

