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Chapter 2: Problem 153

The velocity \(v\) of a body moving along a straight line is varying with time \(t\) as \(v=t^{2}-4 t\), where \(v\) in \(\mathrm{m} / \mathrm{s}\) and \(t\) in seconds. The magnitude of initial acceleration is (A) Zero (B) \(2 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(4 \mathrm{~m} / \mathrm{s}^{2}\) (D) \(6 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The magnitude of the initial acceleration is 4 m/s虏 (Option C).

Step by step solution

01

Finding the acceleration function

To find the acceleration function, we need to differentiate the velocity function with respect to time \(t\). The given velocity function is: \(v(t) = t^2 - 4t\) Now, let's differentiate the velocity function with respect to \(t\): \( a(t) = \frac{d}{dt}(t^2 - 4t)\)
02

Differentiate the velocity function

Using basic differentiation rules, differentiate the velocity function: \( a(t) = \frac{d}{dt}(t^2) - \frac{d}{dt}( 4t)\) \( a(t) = 2t - 4\) Now we have the acceleration function.
03

Finding the initial acceleration

To find the initial acceleration (when \(t = 0\)), substitute \(t = 0\) into the acceleration function: \( a(0) = 2 \cdot 0 - 4\) \( a(0) = -4\) The magnitude of the initial acceleration is the absolute value of the initial acceleration: \(|a(0)| = |-4| = 4\) m/s虏 Therefore, the correct answer is: (C) \(4 \mathrm{~m} / \mathrm{s}^{2}\)

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Key Concepts

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

Kinematics
Kinematics is a branch of classical mechanics that deals with the description of motion without considering the forces that cause it.

At the core of kinematics are the equations of motion, which provide the mathematical relationship between the kinematic quantities: displacement, velocity, acceleration, and time. In the context of the exercise, we focus on velocity, which is the rate of change of displacement with respect to time, and acceleration, which is the rate of change of velocity with respect to time.

Understanding kinematics involves recognizing how these quantities evolve over time for a moving object. As in our problem, the velocity function was given by a polynomial expression of time, indicating that the body's motion is changing as time progresses.
Differentiation in Physics
Differentiation is a mathematical tool used extensively in physics to understand rates of change. In kinematics, differentiation is applied to find acceleration since acceleration is the derivative of velocity with respect to time.

This involves applying the principles of calculus to determine how the velocity of an object changes over an interval of time. For the initial acceleration, we take the derivative of the velocity function at time zero.

Practical Application of Differentiation

The process is not just theoretical; it has practical implications in predicting how quickly an object is going to speed up or slow down. Being able to calculate these rates of change helps in designing safe transportation systems, predicting planetary movements, and even in the precise timing needed in athletic pursuits.
Motion Along a Straight Line
When an object is said to move along a straight line, it implies a one-dimensional motion where only one spatial coordinate changes with time. This is the simplest form of motion studied in kinematics, and it simplifies analysis, as there's no need to consider complex trajectories or multiple components of motion.

In the given exercise, the body's motion along a straight line is fully described by the velocity as a function of time, and subsequent differentiation gives us the acceleration.

Real-World Implications

Understanding straight-line motion is key to solving real-world problems like traffic flow, the launching of projectiles, and even the motion of particles in accelerator experiments. It forms the foundation upon which more elaborate motions are studied and understood.

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Most popular questions from this chapter

The maximum height of a projectile for two complementary angles of projection is \(50 \mathrm{~m}\) and \(30 \mathrm{~m}\) respectively. The initial speed of projectile is (A) \(10 \sqrt{34} \mathrm{~m} / \mathrm{s}\) (B) \(40 \mathrm{~m} / \mathrm{s}\) (C) \(20 \mathrm{~m} / \mathrm{s}\) (D) \(10 \mathrm{~m} / \mathrm{s}\)

Two bodies are projected at angles \(\theta\) and \((90-\theta)\) to the horizontal with the same speed. The ratio of their times of flight is (A) \(\sin \theta: 1\) (B) \(\cos \theta: 1\) (C) \(\sin \theta: \cos \theta\) (D) \(\cos \theta: \sin \theta\)

If \(\vec{r}=b t \hat{i}+c t^{2} \hat{j}\) where \(b\) and \(c\) are positive constants, the velocity vector make an angle of \(45^{\circ}\) with the \(x\) and \(y\) axes at \(t\) equal to (A) \(\frac{b}{2 c}\) (B) \(\frac{b}{c}\) (C) \(\frac{c}{2 b}\) (D) \(\frac{c}{b}\)

A ball is thrown from a point with a speed ' \(v_{0}\) ' at an elevation angle of \(\theta\). From the same point and at the same instant, a person starts running with a constant speed \(\frac{' v_{0}}{2}\) to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection? (A) No (B) Yes, \(30^{\circ}\) (C) Yes, \(60^{\circ}\) (D) Yes, \(45^{\circ}\)

Two boys (at ground) simultaneously aim their guns at a bird sitting on a tower. The first boy releases his shot with speed of \(100 \sqrt{2} \mathrm{~m} / \mathrm{s}\) at an angle \(45^{\circ}\) with the horizontal. The second boy is behind the first boy by a distance \(\quad 100(\sqrt{3}-1) \mathrm{m}\) and releases his shot with speed \(200 \mathrm{~m} / \mathrm{s}\). Both the shots hit the bird simultaneously. Angle of projection of the shot fired by the second boy is (A) \(\frac{\pi}{6}\) (B) \(\frac{\pi}{3}\) (C) \(\frac{\pi}{4}\) (D) None of these
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