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Chapter 2: Problem 196

The distance travelled by a particle is given by \(\mathrm{s}=3+2 \mathrm{t}+5 \mathrm{t}^{2}\) The initial velocity of the particle is \(\ldots\) (A) 2 unit (B) 3 unit (C) 10 unit (D) 5 unit

Short Answer

Expert verified
The initial velocity of the particle is \(s'(0) = 2 + 10(0) = 2\), which is (A) 2 units.

Step by step solution

01

Identify the given function

The distance travelled by the particle as a function of time is given by: s(t) = 3 + 2t + 5t^2
02

Find the first derivative of s(t) with respect to t

Using the power rule, the first derivative of s(t) with respect to t is: s'(t) = \(\frac{d}{dt}(3) + \frac{d}{dt}(2t) + \frac{d}{dt}(5t^2)\)
03

Apply the power rule for each term

The power rule states that \(\frac{d}{dt}(k*t^n)=k*n*t^{n-1}\), where k and n are constants. Applying this rule to each term, we have: s'(t) = \(\frac{d}{dt}(3) + \frac{d}{dt}(2t) + \frac{d}{dt}(5t^2)\) = 0 + 2 + 10t
04

Evaluate s'(t) at t = 0

The initial velocity corresponds to the value of s'(t) when t = 0: s'(0) = 2 + 10(0) = 2
05

Identify the correct option

Comparing the result with the given options, the correct answer is (A) 2 units. So, the initial velocity of the particle is 2 units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Derivatives
In calculus, a derivative represents the rate at which a function is changing at any given point. When tackling problems involving motion, derivatives are critical.
The derivative of a function can give us insights into how the function behaves with respect to its varying inputs. This is particularly helpful in physics and engineering, where motion is often described mathematically.
  • The first derivative of a position function, such as velocity, tells us how fast the position changes, i.e., it provides the speed or velocity of the object.
  • By applying rules such as the power rule, derivatives can be quickly calculated for polynomial functions. The power rule is applied by multiplying the base's power by its coefficient and subtracting one from the power.
In our problem, taking the derivative of the position function with respect to time provides us with the velocity function, which describes how the velocity of the particle changes over time.
Basics of Kinematics
Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause this motion. It is crucial to understand how objects behave as they move over time.
Kinematics primarily involves parameters such as displacement, velocity, and acceleration. Here's a brief overview:
  • Displacement is the object鈥檚 overall change in position.
  • Velocity refers to the rate of change of displacement with time. It has both magnitude and direction, making it a vector quantity.
  • Acceleration is the rate of change of velocity with time, indicating whether the object is speeding up, slowing down, or changing direction.
In our exercise, understanding kinematics helps us determine the initial velocity, which is the velocity at the starting point of time, t = 0. By calculating the first derivative of the distance function with respect to time, we can determine the particle's velocity at any given moment.
Exploring the Velocity-Time Relation
The velocity-time relationship helps us distinguish changes in an object's velocity over time. This relationship can be understood through a velocity-time graph or mathematically by evaluating the derivative.
With time being a continuous variable, velocity can at any point give insights into the nature of the motion:
  • Instantaneous velocity at a specific time is found using the derivative, which we've already calculated in our solution as the function of velocity.
  • When evaluating at initial conditions, such as t = 0, we derive what is known as the initial velocity.
In the given problem, the velocity-time relation helps us assert that the initial velocity is obtained by evaluating the velocity function at zero time. Hence, through proper evaluation of derivatives and the velocity-time relation, the initial velocity of 2 units is accurately determined.

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Most popular questions from this chapter

If \(\mathrm{f}\) is the frequency of a body moving in a circular path with constant speed. a is its centrifugal acceleration, so. (A) \(\mathrm{a} \propto \mathrm{f}\) (B) a \(\propto \mathrm{f}^{2}\) (C) \(\mathrm{a} \propto \mathrm{f}^{3}\) (D) a \(\propto(1 / \mathrm{f})\)

Magnitudes of \(\mathrm{A}^{\rightarrow}, \mathrm{B}^{\rightarrow}\) and \(\mathrm{C}^{\rightarrow}\) are 41,40 and 9 respectively, \(\mathrm{A}^{\rightarrow}=\mathrm{B}^{\rightarrow}+\mathrm{C}^{\rightarrow}\) Find the angle between \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) (A) \(\sin ^{-1}(9 / 40)\) (B) \(\cos ^{-1}(9 / 41)\) (C) \(\tan ^{-1}(9 / 41)\) (D) \(\tan ^{-1}(41 / 40)\)

An object moves in a straight line. It starts from the rest and its acceleration is \(2 \mathrm{~ms}^{-2}\). After reaching a certain point it comes back to the original point. In this movement its acceleration is \(-3 \mathrm{~ms}^{-2}\). till it comes to rest. The total time taken for the movement is 5 second. Calculate the maximum velocity. (A) \(6 \mathrm{~ms}^{-1}\) (B) \(5 \mathrm{~ms}^{-1}\) (C) \(10 \mathrm{~ms}^{-1}\) (D) \(4 \mathrm{~ms}^{-1}\)

Comprehensions type questions. A particle is moving in a circle of radius \(\mathrm{R}\) with constant speed. The time period of the particle is T Now after time \(\mathrm{t}=(\mathrm{T} / 6)\) Average speed of the particle is (A) \((\pi \mathrm{R} / 6 \mathrm{~T})\) (B) \([(2 \pi R) / 3 \mathrm{~T}]\) (C) \([(2 \pi R) / T]\) (D) \((\mathrm{R} / \mathrm{T})\)

A particle moves in straight line. Its position is given by \(\mathrm{x}=2+5 \mathrm{t}-3 \mathrm{t}^{2}\). Find the ratio of initial velocity and initial acceleration. \((\mathrm{A})+(5 / 6)\) (B) \(-(5 / 6)\) (C) \((6 / 5)\) (D) \(-(6 / 5)\)
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