Question

For a particle moving along x axis velocity time graph as shown in the figure. find the distance traveled and displacement of the particle.?​

Answer 1

Answer:

Check out the attachment...

Answer 2

Concept:

When an object is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.

A displacement is a vector whose length is the shortest distance between a moving object's starting and final positions.

Given:

The velocity-time graph along the x-axis.

Find:

The distance traveled and the displacement of the particle.

Solution:

The distance traveled by the particle can be calculated by adding the area of the trapezium and the area of the triangle.

Distance traveled = Area of ABG + Area of BCHG + Area of CHJ + Area of JDE

$d=\frac{1}{2}(2)(8)+(2)(8)+\frac{1}{2} (2)(8)+\frac{1}{2}(4)(5)$

$d=8+16+8+10=42m$

The displacement of the particle is the area under the velocity time graph.

Therefore,

displacement $=\frac{1}{2} (2)(8)+(2)(8)+\frac{1}{2} (2)(8)+[\frac{1}{2} (4)(-5)$

displacement $=32m-10m=22m$

The distance traveled and the displacement of the particle is $42m$ and $22m$ respectively.

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