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Chapter 5: Problem 1

A particle is moving in xy-plane with \(y=x / 2\) and \(v_{x}=4-2 t\). Choose the correct options. (1) Initial velocities in \(x\) and \(y\) directions are negative. (2) Initial velocities in \(x\) and \(y\) directions are positive. (3) Motion is first retarded, then accelerated. (4) Motion is first accelerated, then retarded.

Short Answer

Expert verified
(2) Initial velocities in x and y directions are positive. (3) Motion is first retarded, then accelerated.

Step by step solution

01

Determine the Initial Velocity in the x Direction

The velocity in the x-direction is given by the function \(v_{x} = 4 - 2t\). To find the initial velocity, substitute \(t = 0\):\[v_{x}(0) = 4 - 2(0) = 4.\]The initial velocity in the x-direction is 4, which is positive.
02

Determine the Initial Velocity in the y Direction

Since \(y = \frac{x}{2}\), differentiation with respect to time gives us the velocity in the y-direction:\[v_{y} = \frac{dx}{dt} \cdot \frac{1}{2} = \frac{v_{x}}{2} = \frac{4 - 2t}{2} = 2 - t.\]Find the initial velocity in the y-direction at \(t = 0\):\[v_{y}(0) = 2 - 0 = 2.\]The initial velocity in the y-direction is 2, which is positive.
03

Analyze the Nature of Motion

The velocity in the x-direction, \(v_{x} = 4 - 2t\), decreases with time since it has a negative coefficient with \(t\). The motion in the x-direction is retarded initially.For \(v_{y} = 2 - t\), velocity in the y-direction also decreases with time. Therefore, the motion is retarded initially in both directions.As \(t\) increases and \(v_{x}\) and \(v_{y}\) become eventually zero and then negative, acceleration in this direction will also switch.
04

Motion Characteristics Over Time

For \(t < 2\), both \(v_{x}\) and \(v_{y}\) are positive (motion in positive x and y directions), but decreasing. When \(t = 2\), the velocities become zero, changing the state of motion from retardation to acceleration in the opposite direction.After \(t=2\), both \(v_{x}\) and \(v_{y}\) become negative but increasing (magnitude wise), indicating acceleration.

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Key Concepts

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

Velocity in x-direction
The velocity in the x-direction specifies how quickly the particle is moving horizontally along the x-axis. In this particular problem, the velocity in the x-direction is described by the equation \(v_{x} = 4 - 2t\). This means that the initial velocity when time \(t\) is zero is positive and equal to 4 units.

Here are important details about the velocity in the x-direction:
  • At \(t = 0\), \(v_{x}(0) = 4\), indicating the initial horizontal movement is in the positive direction.
  • As time progresses, the term \(-2t\) causes the velocity to decrease over time. This decrease implies a reduction in speed, which we recognize as retardation in this context.
Eventually, when \(t = 2\), \(v_{x}\) reaches zero, and continues to decrease, turning negative for \(t > 2\). When \(v_{x}\) becomes negative, the particle reverses and moves in the opposite direction but starts accelerating.
Velocity in y-direction
Parallel to our exploration of horizontal movement, let's look at the velocity in the y-direction, controlling the vertical motion. Starting from the given equation \(y = \frac{x}{2}\), deriving with respect to time gives us the velocity in y-direction: \(v_{y} = \frac{4 - 2t}{2} = 2 - t\).

Here are the aspects to comprehend about \(v_{y}\):
  • Initially, at \(t = 0\), the velocity in the y-direction is \(2\), indicating vertical movement upwards.
  • Similar to \(v_{x}\), \(v_{y}\) too decreases over time because of the \(-t\) term in the equation, leading to retardation.
When time \(t\) reaches 2, \(v_{y}\) is zero, marking that upward motion ceases, and the velocity then starts turning negative, showing that the particle begins moving in the opposite vertical direction, accelerating as it moves downward.
Retardation and Acceleration
Retardation and acceleration, essential concepts in motion dynamics, describe changes in velocity. Retardation, also known as negative acceleration, implies a decrease in speed. Acceleration, conversely, is an increase in speed, which can either increase or decrease depending on the direction of motion.

In this scenario, the changes work as follows:
  • Initially, both \(v_{x}\) and \(v_{y}\) decrease in magnitude due to the negative coefficients, indicating retardation. Both directions witness a slowing down of motion as \(t\) progresses.
  • At \(t = 2\), the velocities drop to zero, pausing any further motion but setting the stage for directional change.
  • Beyond \(t = 2\), the previously retarded motion becomes acceleration, as both velocity components \(v_{x}\) and \(v_{y}\) switch their signs, and the particle begins to speed up again, but in the opposite direction.
Understanding the transition from retardation to acceleration is key to mastering motion in two dimensions, highlighting how velocity components interact with time to alter movement directions.

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Most popular questions from this chapter

A particle is fired from \(A\) in the diagonal plane of a building of dimension \(20 \mathrm{~m}\) (length) \(\times 15 \mathrm{~m}\) (breadth) \(\times 12.5\) (height). just clears the roof diagonally and falls on the other side of the building at \(B\). It is observed that the particle is travelling at an angle \(45^{\circ}\) with the horizontal when it clears the edges \(P\) and \(Q\) of the diagonal. Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) The angle of projection at \(A\) will be: (1) \(30^{\circ}\) (2) \(45^{\circ}\) (3) \(60^{\circ}\) (4) \(75^{\circ}\)

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The speed of a projectile at its highest point is \(v_{1}\) and at the point half the maximum height is \(v_{2}\). If \(\frac{v_{1}}{v_{2}}=\sqrt{\frac{2}{5}}\), then fin? the angle of projection. (1) \(45^{\circ}\) (2) \(30^{\circ}\) (3) \(37^{\circ}\) (4) \(60^{\circ}\)

We know that when a boat travels in water, its net velocity w.r.t. ground is the vector sum of two velocities. First is the velocity of boat itself in river and other is the velocity of water w.r.t. ground. Mathematically: \(\vec{v}_{\text {boat }}=\vec{v}_{\text {boat, water }}+\vec{v}_{\text {water }}\) Now given that velocity of water w.r.t. ground in a river is \(u\). Width of the river is \(d\). A boat starting from rest aims perpendicular to the river with an acceleration of \(a=5 t\), where \(t\) is time. The boat starts from point \((1,0)\) of the coordinate system as shown in figure. Assume SI units. Obtain the total time taken to cross the river. (1) \((3 d / 5)^{1 / 3}\) (2) \((6 d / 5)^{1 / 3}\) (3) \((6 d / 5)^{1 / 2}\) (4) \((2 d / 3)^{1 / 3}\)
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