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The position of a particle along \(x\) -axis at time \(t\) is given by \(x=1+t-t^{2}\). The distance travelled by the particle in first 2 seconds is (a) \(1 \mathrm{~m}\) (b) \(2 \mathrm{~m}\) (c) \(2.5 \mathrm{~m}\) (d) \(3 \mathrm{~m}\)

Short Answer

Expert verified
The total distance travelled by the particle in the first 2 seconds is 2.5 m.

Step by step solution

01

Determine the Initial and Final Positions

The position function is given by \(x(t) = 1+t-t^{2}\). First, find the initial position at \(t=0\) by substituting 0 into the function: \(x(0) = 1+0-0^{2} = 1\). Next, find the position at \(t=2\): \(x(2) = 1+2-2^{2} = 1+2-4 = -1\).
02

Use the Position Values to Determine Total Distance

The particle starts at \(x=1\) at \(t=0\) and ends at \(x=-1\) at \(t=2\). Direct subtraction to find distance is not enough due to the potential change in direction. The key is checking the vertex of the parabolic journey.
03

Find the Vertex of the Parabola

The motion equation \(x(t) = 1 + t - t^2\) is a downward open parabola. The vertex is crucial as it signifies change in direction. For \(x = at^2 + bt + c\), the vertex occurs at \(t = -\frac{b}{2a}\). Here, \(a = -1\), \(b = 1\), hence the vertex is at \(t = \frac{1}{2}\).
04

Evaluate Positions at Key Time Intervals

Calculate the position at the vertex (\(t = 0.5\)): \(x(0.5) = 1 + 0.5 - (0.5)^2 = 1 + 0.5 - 0.25 = 1.25\). Hence, the particle moved from \(x = 1\) to \(x = 1.25\), and then from \(x = 1.25\) to \(x = -1\).
05

Calculate Total Distance Travelled

The distance travelled from \(t=0\) to \(t=0.5\) is \(|1.25 - 1| = 0.25 \text{ m}\). From \(t=0.5\) to \(t=2\) is \(|-1 - 1.25| = 2.25 \text{ m}\). Thus, the total distance travelled is \(0.25 + 2.25 = 2.5 \text{ m}\).

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Key Concepts

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

Position-Time Relation
The position-time relation is a fundamental concept in kinematics that describes how the position of a particle changes over time. This relation is important for analyzing motion along a line, such as the motion of a particle along the x-axis.
The position of a particle is often described using a function of time, such as in the equation given by the exercise: \(x(t) = 1 + t - t^2\). This equation gives the position \(x\) of the particle at any time \(t\). By substituting different values of \(t\) into the equation, we can determine where the particle is located at those times.
In the exercise, we calculated the initial position \(x(0)\) and final position \(x(2)\) to understand the particle's journey over a particular time interval.
Parabolic Motion
Parabolic motion refers to any motion following a path that forms a parabola. In this exercise, the function describing the particle's movement \(x(t) = 1 + t - t^2\) is a quadratic equation, indicating that the motion path is a parabola.
Parabolas open either upwards or downwards depending on the sign of the quadratic term (\(t^2\)). Here, the parabola opens downwards because the coefficient of \(t^2\) is negative (\(-1\)). Understanding the nature of the parabola gives insight into the change of direction of the particle. The particle will reach its maximum position at the vertex, and then reverse direction.
This allows us to predict the general motion and anticipate key points in the particle's journey, such as turning points or maximum height.
Distance Calculation
Calculating the distance travelled by a particle involves evaluating the total length of the path it covers, regardless of changes in direction. It is crucial to consider every segment of the journey. This is particularly important when dealing with parabolic motion, where a simple calculation by subtracting final and initial positions would not suffice.
For example, in finding the distance traveled between \(t = 0\) and \(t = 2\), we need to account for the vertex, where the direction changes. We calculate the partial distances from \(x = 1\) to the vertex \(x = 1.25\) and from the vertex to the final position \(x = -1\).
Adding these segments together yields the total distance: \(0.25\) m from start to vertex and \(2.25\) m from vertex to end, totaling \(2.5\) m.
Vertex of a Parabola
In the context of motion, the vertex of a parabola is a critical point. It represents the peak or trough of the trajectory, where the object changes direction. For the equation \(x(t) = 1 + t - t^2\), finding the vertex helps us understand exactly when and where this change occurs.
To find the vertex in a parabolic equation \(at^2 + bt + c\), use the formula \(t = -\frac{b}{2a}\). Applying this to our equation with \(a = -1\) and \(b = 1\), we calculate \(t = \frac{1}{2}\). At this point, we plug it back into the position function to find \(x(0.5) = 1.25\).
Knowing the vertex allows us to accurately determine the distance the particle travels before and after reaching this peak, ensuring precise distance calculations.

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Most popular questions from this chapter

'lhe velocity of a particle is zero at \(t=0\). 'Ihen, (a) the acceleration at \(t=0\) must be zero (b) if acceleration is zero from \(t=0\) to \(t=10 \mathrm{~s}\), the speed is also zero in this interval (c) if speed is zero from \(t=0\) to \(t=10 \mathrm{~s}\), the acceleration is also zero in this interval (d) if acceleration is zero from \(t=0\) to \(t=10 \mathrm{~s}\), the displacement is also zero in this interval

Statement-1: A body having non-zero acceleration can have a constant velocity. Statement-2 : Acceleration is the rate of change of velocity.

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