Question

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) \(4 \mathrm{~ms}^{-1}\)

Short answer

The maximum velocity is \(6 \mathrm{~ms}^{-1}\) (A).

Step-by-step solution

Write equations for displacement for each part of motion

Since the object comes back to the original position, the displacement in each part of the motion must be equal and opposite. Using the equation: \(x = ut + \frac{1}{2}at^2\), we have: For the first part of the motion: \(x = 0 + \frac{1}{2}(2)t_1^2\) For the second part of the motion: \(-x = 0 + \frac{1}{2}(-3)t_2^2\)

Solve for the time taken in each part of the motion

Using the given information t1 + t2 = 5 s, and equating the magnitudes of displacement from the previous step, we have: \(t_1^2 = \frac{3}{2}t_2^2\) Substituting t2 = 5 - t1, we get: \(t_1^2 = \frac{3}{2}(5 - t_1)^2\) Solve for t1: \(t_1 \approx 3 \mathrm{s}\) So, t2 = 5 - t1 = 2 s.

Find maximum velocity

Now we know the time taken for the first part of the motion, we can calculate the maximum velocity using the initial velocity and acceleration for that part: \(v_\text{max} = u_1 + a_1t_1\) Since the object starts from rest, u1 = 0. The acceleration for the first part of the motion, a1 = 2 ms^-2, and the time taken, t1 = 3 s. Thus, we can plug in these values to find the maximum velocity: \(v_\text{max} = 0 + (2)(3)\) \(v_\text{max} = 6 \mathrm{~ms}^{-1}\) Therefore, the maximum velocity is \(6 \mathrm{~ms}^{-1}\) and the correct answer is (A) \(6 \mathrm{~ms}^{-1}\).

Key concepts

Acceleration

In the study of kinematics, acceleration is a crucial concept. It is defined as the rate at which an object changes its velocity. When an object speeds up, slows down, or changes direction, it is accelerating. Acceleration can have positive or negative values, where positive indicates an increase in velocity and negative indicates a decrease.

In the given exercise, the object experiences two phases of acceleration:

• The first phase has a constant acceleration of \(2 \mathrm{~m/s^2}\).
• The second phase has an acceleration of \(-3 \mathrm{~m/s^2}\), indicating it's slowing down as it returns to the starting point.

Understanding acceleration is important for calculating changes in velocity and displacement, making it a fundamental piece in solving motion-related problems in physics.

Displacement

Displacement refers to the overall change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. Unlike distance, which only considers the length of the path traveled, displacement considers the shortest path and direction from the starting point to the ending point.

In this exercise, the object returns to its original position, resulting in a net displacement of zero. During the first phase, the object moves in a positive direction, and during the second phase, it moves in the negative direction with equal magnitude, leading to no net change in position by the end of the motion. This property of displacement being zero plays a crucial role in solving the problem as it hints at the equality of distances covered in both phases.

Velocity

Velocity is another key concept in motion, describing how fast an object is moving and in which direction. It is a vector quantity with both magnitude (speed) and direction. Calculating velocity involves considering both displacement and the time over which the movement occurs.

For the given problem, we were tasked to find the maximum velocity achieved by the object. Using the formula for velocity, \(v = u + at\), and knowing the object starts from rest (initial velocity \(u = 0\)) and accelerates at \(2 \mathrm{~m/s^2}\) for a time \(t_1\), we calculated the maximum velocity as \(6 \mathrm{~m/s}\). This calculation underlines the importance of understanding initial conditions, such as starting from rest, and how they influence velocity.

Equations of motion

Equations of motion are mathematical formulas used to relate displacement, velocity, acceleration, and time. They are essential tools in solving kinematic problems, providing a systematic way to derive unknown variables given known quantities.

In solving the problem, the equation \(x = ut + \frac{1}{2}at^2\) was used to describe the motion during each phase. These equations helped to set up the relationship between the two time intervals \(t_1\) and \(t_2\) by equating their magnitudes of displacement.

This approach allowed us to utilize another fundamental equation, \(v = u + at\), to determine the maximum velocity, showcasing the interconnectedness of the motion parameters.

• The first equation provided us the displacement conditions.
• The second calculated the velocity using the known acceleration and time.

Understanding these equations makes tackling kinematic problems much simpler and intuitive.