Question

A particle moves 8m along a straight line .Then it turns left and moves 6m before finally coming to rest .The total time elapsed is 20s .What is the magnitude of average velocity of the particle​

Answer 1

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Answer 2

Answer:

The magnitude of average velocity of the particle = 0.5

Explanation:

Average Velocity:

An average velocity is defined as the change in position or displacement divided by the time interval in which the displacement occurs.

$Average \ velocity = \frac{Change \ in \ position/Displacement}{Time \ taken}$

                        $\bar v = \frac{\triangle x}{\triangle t}$

It is a vector quantity. The SI unit of average velocity is m/s or ms⁻¹

Given:

The particle moves 8m along a straight line, then turns left and moves 6m.

Total time taken by the particle = 20 sec

Find:

Magnitude of average velocity = ?

Solution:

Refer to the figure given below.

In ΔABC,

AB = 8 m

BC = 6 m

Here, the distance covered by the particle = AB + BC

                                                                 = $8+6$

                                                                 = $14 \ m$

Displacement of the particle = AC = $x$

Using Pythagoras theorem,

hypotenuse² = base² + altitude²

∴ AC² = AB² + BC²

⇒ $x^{2} = 8^{2} + 6^{2}$

⇒ $x^{2} = 64 +36$

⇒ $x^{2} =100$

⇒ $x=\sqrt{100}$

⇒ $x =10$

∴ Displacement of the particle = 10 m

Now, average velocity = $\frac{Displacement }{Time \ taken}$

                                      = $\frac{10}{20}$

                                      = $0.5\ m/s$

Therefore, the magnitude of average velocity = 0.5

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