Chapter 2: Problem 241
A particle moves in one direction with acceleration \(2 \mathrm{~ms}^{-2}\) and initial velocity \(3 \mathrm{~ms}^{-1}\).After what time its displacement will be \(10 \mathrm{~m}\) ? (A) \(1 \mathrm{~s}\) (B) \(2 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(4 \mathrm{~s}\)
Short Answer
Expert verified
(B) \(2 \mathrm{~s}\)
Step by step solution
01
Write down the given values
We have the following values given:
Acceleration (a) = \(2 \mathrm{~ms}^{-2}\)
Initial velocity (u) = \(3 \mathrm{~ms}^{-1}\)
Displacement (s) = \(10 \mathrm{~m}\)
02
Use the equation \(s = ut + \frac{1}{2}at^2\) to solve for time (t)
Plugging in the given values, we get:
\(10 = 3t + \frac{1}{2}(2)t^2\)
Simplifying the equation:
\(10 = 3t + t^2\)
03
Rearrange the equation to find t
Move the terms to one side:
\(t^2 + 3t - 10 = 0\)
Now, we need to solve this quadratic equation for t.
04
Solve the quadratic equation for t
We can solve this equation by factoring:
\((t+5)(t-2) = 0\)
Therefore, either t = -5 or t = 2. However, time cannot be negative, so we can dismiss the -5 as a solution.
Thus, t = \(2 \mathrm{~s}\).
Our answer is (B) \(2 \mathrm{~s}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acceleration
Acceleration is a fundamental concept in kinematics, describing how quickly an object's velocity changes over time. It is expressed in meters per square second (m/s²). In this exercise, the given acceleration is 2 m/s². This means that each second, the velocity of the particle increases by 2 meters per second.
Understanding acceleration can help predict how an object moves over time under the influence of force.
Understanding acceleration can help predict how an object moves over time under the influence of force.
- Positive acceleration indicates an increase in velocity.
- Negative acceleration, or deceleration, indicates a decrease.
Initial Velocity
Initial velocity is the speed and direction at which an object starts its motion. It serves as the starting point for further calculations involving acceleration and displacement.
In our exercise, the initial velocity (denoted as \(u\)) is given as 3 m/s. This means when the time () count begins, the particle is already moving at 3 meters per second.
Initial velocity plays a significant role when calculating future positions and speeds of an object. Its value is one of the primary parameters in the equation of motion \(s = ut + \frac{1}{2}at^2\). This equation allows you to plug in the initial velocity along with the acceleration and displacement to find unknown variables such as time. Knowing the initial velocity helps you accurately predict how far an object will travel over time.
In our exercise, the initial velocity (denoted as \(u\)) is given as 3 m/s. This means when the time () count begins, the particle is already moving at 3 meters per second.
Initial velocity plays a significant role when calculating future positions and speeds of an object. Its value is one of the primary parameters in the equation of motion \(s = ut + \frac{1}{2}at^2\). This equation allows you to plug in the initial velocity along with the acceleration and displacement to find unknown variables such as time. Knowing the initial velocity helps you accurately predict how far an object will travel over time.
Displacement
Displacement is a vector quantity that represents the change in position of an object. Unlike distance, which covers the entire path traveled, displacement directly measures the shortest path from start to finish with direction.
The goal in the exercise was to find out after how long the particle would reach a displacement of 10 meters. This indicates the particle's final position, accounting for both the initial velocity and the effect of acceleration over time.
Displacement is central to solving kinematic equations because it reflects the net change in position, integrating factors like direction and distance. In the case of this problem, knowing the displacement allows us to use the fundamental equation of motion to solve for time, considering the particle travels in a single direction.
The goal in the exercise was to find out after how long the particle would reach a displacement of 10 meters. This indicates the particle's final position, accounting for both the initial velocity and the effect of acceleration over time.
Displacement is central to solving kinematic equations because it reflects the net change in position, integrating factors like direction and distance. In the case of this problem, knowing the displacement allows us to use the fundamental equation of motion to solve for time, considering the particle travels in a single direction.
Quadratic Equations
Quadratic equations frequently arise in the analysis of motion under constant acceleration. These equations take the form \(ax^2 + bx + c = 0\) and have solutions that may be real or complex numbers.
In this exercise, the equation derived from the motion formula was \(t^2 + 3t - 10 = 0\), where \(t\) denotes time. Solving this quadratic equation informed us about the possible times at which the particle could have achieved the given displacement.
There are multiple ways to solve quadratic equations:
In this exercise, the equation derived from the motion formula was \(t^2 + 3t - 10 = 0\), where \(t\) denotes time. Solving this quadratic equation informed us about the possible times at which the particle could have achieved the given displacement.
There are multiple ways to solve quadratic equations:
- Factoring: If the equation easily breaks down into the product of binomials.
- The Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- Completing the Square: Rewriting the equation to make it easier to resolve.