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Chapter 2: Problem 6

Velocity of a particle is in negative direction with constant acceleration in positive direction. Column - I Column - II (a) Vclocity-time graph (p) Slopc - negative (b) Acceleration-time graph (q) Slopc-positive (c) Displacement-time graph (r) Slopc-zero (s) | Slope \(\mid\) - increasing (1) | Slope \(\mid\) - decreasing (u) | Slope \(\mid\) - constant

Short Answer

Expert verified
(a) matches (q), (b) matches (r), (c) matches (s).

Step by step solution

01

Analyze the velocity-time graph

In a velocity-time graph, if the velocity is in the negative direction and the acceleration is constant in the positive direction, the graph will be a straight line with a positive slope because the velocity increases over time due to positive acceleration. Hence, column I (a) matches column II (q) slope-positive.
02

Analyze the acceleration-time graph

Since the acceleration is constant, the acceleration-time graph will be a horizontal line, reflecting no change in acceleration over time. Therefore, column I (b) corresponds to column II (r) slope-zero.
03

Analyze the displacement-time graph

Initially, with a negative velocity, the particle moves in a negative direction, but as the velocity becomes less negative and eventually positive due to constant positive acceleration, the displacement graph's slope will start increasing (indicating a curve that becomes less steep in the negative direction and starts to move in the positive direction). Therefore, column I (c) matches column II (s) | Slope | - increasing.

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Key Concepts

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

Acceleration-Time Graph
An acceleration-time graph is a helpful tool to visualize how acceleration behaves over time. Imagine this graph as a straight road where you keep the speed of a car exactly constant. This means, no matter how far you drive, your speed stays the same. Likewise, if acceleration stays constant, the graph shows a horizontal line.

In the exercise, we know the acceleration is constant. It does not vary, which is why the acceleration-time graph depicts a flat, horizontal line. This horizontal line indicates a "slope-zero."
  • Constant acceleration: straight horizontal line
  • Slope-zero: indicates no change
Displacement-Time Graph
Understanding a displacement-time graph can help you see how an object's position changes over time. Think of this as watching a runner on a track. As they run, you see them move along the track, which corresponds to displacement over time.

In our context, the particle starts with a negative velocity, almost like a runner starting backward. As positive acceleration works its magic, the particle slows its backward movement and eventually starts moving forward. This change is visible in the displacement-time graph as a curve whose slope initially appears steep but gradually increases and flattens as speed stabilizes.

Hence, this graph鈥檚 slope is said to be "increasing":
  • Negative direction initially: runner starts backward
  • Positive acceleration: curves begin to rise
  • Increasing slope: indicates growing forward motion
Positive Acceleration
Positive acceleration implies that an object is speeding up in the positive direction. Think of this as pressing the accelerator in a car; your speed increases. In physics terms, positive acceleration means the force exerted on an object enhances its velocity along its given path.

For the particle in our example, even though it starts with a negative velocity, the constant positive acceleration gradually changes this by increasing the speed in the positive aspect of the direction. This is why a velocity-time graph with positive acceleration depicts a line with "positive slope."

  • Positive acceleration: velocity increase in positive direction
  • Car example: "pressing the accelerator"
  • Positive slope: line slopes upwards over time
Negative Velocity
Negative velocity describes a scenario where an object moves in the opposite direction to the positive direction defined in a problem. Picture a car reversing; as it moves backward, it covers distance in the negative scope.

Initially, our particle has a negative velocity, implying it's moving in a reverse or backward direction. But with constant positive acceleration acting upon it, this negative value shrinks, becomes zero, and eventually turns positive, indicating forward movement. The concept helps illustrate why on a graph, an initial negative slope begins to increase positively as acceleration takes effect.

  • Negative velocity: reverse or backward movement
  • Positive acceleration effect: changes direction over time
  • Starting slope negative: becomes positive eventually

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Most popular questions from this chapter

Statement-1: The \(v-t\) graph perpendicular to tho time axis not possible. Statement-2 : If \(v-t\) graph is perpendicular to the time axis, then velocity of particle should be infinite.

'Two particles start moving from the same point along the same straight line. The first moves with constant velocity \(v\) and the second with constant acceleration \(a\). During the time that elapses before the second catches the first, what is the greatest distance between the particles.

A marble ball, after having fallen from rest undor the influenee of gravity for \(6 \mathrm{sec}\). erashes through a horizontal glass plate, thercby losing two-third of its velocity. If it then reaches the ground in \(2 \mathrm{sec}\), find the height of the plate above the ground

From the top of a tower of height \(200 \mathrm{~m}\), a ball \(A\) is projected up with \(10 \mathrm{~ms}^{1}\) and two seconds later another ball \(B\) is projected vertically down with the same speed. Then, (a) both \(A\) and \(B\) will reach the ground simultaneously. (b) the ball \(A\) will hit the ground 2 second later than \(B\) hitting the ground. (c) both the balls will hit the ground with the same velocity. (d) both will rebound to the same height from the ground, if both have same coefficient of restitution.

Starting from rest a particle moves in a straight line with acceleration \(a=\left(25-t^{2}\right)^{1 / 2} \mathrm{~m} / \mathrm{s}^{2} \quad\) for \(0 \leq t \leq 5 \mathrm{~s}\) \(a=\frac{3 \pi}{8} \mathrm{~m} / \mathrm{s}^{2} \quad\) for \(t>5 \mathrm{~s}\) What is the velocity of particle at \(t=7 \mathrm{~s}^{\text {? }}\)
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