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In Physics, Motion is the state of body in which it changes its position with time. Motion is fundamentally described by physical quantities such distance, displacement, speed, velocity, acceleration, and time. These physical quantities can be expressed in the form of a mathematical equation to express motion. These equations are called Equations of Motion. These equations can be derived via various methods such as Algebraic Method, Graphical Method and Calculus Method. This article deals with the equation of motion and its derivation using the calculus method. This derivation is useful for class 11 students. Table of Content Fundamentals of MotionThe fundamentals of motion in physics are described by the concepts: of distance, displacement, speed, velocity, acceleration, and time. Here’s a brief overview: Distance and Displacement
Speed and Velocity
Acceleration
Time
Equations of Motion by Calculus MethodThere are three equations of motions also called Newton’s Equation of Motion. The expression of the three equations of motions are:
In the above three equations, v is final velocity, u is initial velocity, a is acceleration, s is distance and t is time. Now, let’s derive these equation by calculus method: First Equation of Motion by CalculusNewton’s first equation of motion states that final velocity is equal to sum of initial velocity and product of acceleration and time. Mathematically, first equation is expressed as v = u + at Mathematically, acceleration is defined as a = dv / dt Integrating both sides of the equation with respect to time gives us: v∫u dv= t∫0 a dt Where v is the final velocity, u is the initial velocity, a is the constant acceleration, and t is the time. ⇒ v∫u dv= at∫0 dt ⇒ v – u = at Adding u on both sides, we get the final form
Second Equation of Motion by CalculusNewton’s second equation of motion states that total distance covered is equal to sum of product of initial velocity and time and half of product of acceleration and square of time. Second equation of motion is mathematically expressed as s = ut + 1/2at2 Now, velocity is rate of change of distance. Hence, in terms of differentiation it can be expressed as v = ds / dt Integrating both sides of the equation with respect to time gives us: s∫0 ds= t∫0 v dt Now using the first equation of motion s∫0 ds= t∫0 (u+ at) dt Where v is the final velocity, u is the initial velocity, s is the displacement, a is the constant acceleration, and t is the time. s∫0 ds = t∫0 u dt + t∫0 at dt Since, u and a are constant ⇒ s∫0 ds = ut∫0 dt + at∫0 t dt ⇒ s = ut + at2/2 Hence, we derive the second equation of motion
Third Equation of Motion by CalculusThird equation of motion states that difference between square of final velocity and initial velocity is equal to the twice the product of acceleration and distance. Mathematically, third equation of motion is given as: v2 – u2 = 2as We start with the mathematical definition of acceleration a = dv/dt Multiply with v on both sides av = v dv/dt Substituting the definition of v = ds/dt a ds/dt = v dv/dt The time dependence can be removed here and this becomes a ds = v dv Integrating both sides s∫0 a ds= v∫u v dv Where v is the final velocity, u is the initial velocity, s is the displacement, and a is the constant acceleration. as = [v2/2]uv (v2 – u2)/2 = as ⇒ v2 – u2 = 2as Rearranging, we get the Third Equation of Motion
Applications of Calculus in MotionHere are some applications of calculus in motion: Analysing Uniformly Accelerated MotionCalculus is used to derive the equations of motion for uniformly accelerated motion. It also enables us to determine instantaneous velocity and acceleration at any given point in time during the motion. By taking derivatives of displacement with respect to time, we obtain velocity functions. Similarly, taking derivatives of velocity with respect to time gives acceleration functions. Calculus allows us to understand how velocity and acceleration change over time. It helps in determining maximum velocity, maximum displacement, and time taken to reach specific positions. Calculating Displacement and Velocity in Variable AccelerationsIn cases of variable acceleration, calculus plays a crucial role in calculating displacement and velocity accurately. The velocity of the object at any time t is the integral of its acceleration function from the initial time t0 to t: v(t) = t∫t0 a(t)dt The displacement of the object from the initial position at time t is the integral of its velocity function from the initial time t0 to t: s(t) = t∫t0 v(t)dt Next step is to evaluate the integrals, and apply initial conditions to determine constants of integration. This would give the final form required.
Examples on Equation of Motion by Calculus MethodExample 1: A body moving along the x-axis has its position given by x(t)=t3-2t+1. Find its velocity and acceleration as a function of time. Solution:
Example 2: For the above problem find the velocity and acceleration at t = 3s. Solution:
Example 3: A body moving along the x-axis has its velocity given by v(t) = t3-2t+1. Find its displacement between t = 0 to 5s. Solution:
Example 4: A body has acceleration varying as a(t) = 6-t. Find its displacement in t = 0 → 3.5s Solution:
Practice Problems on Equation of Motion by Calculus MethodQ1. A car moves along a straight road with its position given by s(t)=2t2-5t+3, where s is in metres and t is in seconds. Determine the car’s velocity and acceleration at time t =3s. Q2. The height of a ball thrown upwards is given by h(t)=3.5t-4.9t2, find its velocity and acceleration as a function of time. Also, find the maximum height reached by the ball. Q3. A particle is moving along the y-axis, and its velocity is given by v(t)= 2t2-3t+5 m/s. Find its displacement between t =1 second and t =4 seconds. Q4. An object is dropped from a height of 100 metres. Its velocity function as it falls is given by v(t)=9.8t, where t is the time in seconds and v is in metres per second. Determine the height of the object after the first 5 seconds of its fall. Q5. A car accelerates with an acceleration function a(t)=2t2−3 m/s². Calculate the displacement of the car from t=0 to t=6 seconds. Conclusion: Equation of Motion by Calculus MethodCalculus plays a vital role in motion analysis by enabling the derivation of Newton’s equations of motion. Additionally, it provides the means to address scenarios of variable accelerations or variable velocity. Calculus allows Newton’s equations to be used for non-linear motion or variable acceleration. FAQs on Equation of Motion by Calculus MethodHow many Equations of Motion are there?
What are other ways to derive equation of motion?
Why is Calculus Important in the Study of Motion?
Can Newton’s Equations Be Used for Non-Linear Motion?
What are the limitations of classical equations of motion?
How to express velocity in differential form?
How to express acceleration in differential form?
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Class 11 |
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