Question

A particle moves along X-axis such that its position with respect to the origin is given  by x = bt2, where b is 0.4 m/s2. Its average velocity during the time interval t = 1 s to t = 5 s is​

Answer 1

Answer:

$\boxed{\mathfrak{Average \ velocity \ (v_{avg}) = 2.4 \ m/s}}$

Given:

Position of a particle w.r.t origin:

x = bt²

where b = 0.4 m/s²

To Find:

Average velocity ($\sf v_{avg}$) during the time interval t = 1s to t = 5s

Explanation:

Average velocity is defined as the change in position or displacement of particle divided by the time interval in which the displacement occurs.

$\boxed{ \bold{v_{avg} = \frac{Change \ in \ position}{Total \ time \ taken} = \frac{x_2 - x_1}{t_2 - t_1}}}$

For the given question:

$\sf t_2 = 5 s$ & $\sf t_1 = 1 s$

$\sf x_2 = x_{t = 5}$ & $\sf x_1 = x_{t = 1}$

$\sf \implies v_{avg} = \frac{x_{t = 5} - x_{t = 1}}{5 - 1} \\ \\ \sf \implies v_{avg} = \frac{b {(5)}^{2} - b {(1)}^{2} }{5 - 1} \\ \\ \sf \implies v_{avg} = \frac{b( {5}^{2} - {1}^{2} )}{4} \\ \\ \sf \implies v_{avg} = \frac{0.4(25 - 1)}{4} \\ \\ \sf \implies v_{avg} = \frac{0.4 \times 24}{4} \\ \\ \sf \implies v_{avg} =0.1 \times 24 \\ \\ \sf \implies v_{avg} =2.4 \: m/s$

$\therefore$

Average velocity ($\sf v_{avg}$) during the time interval t = 1s to t = 5s = 2.4 m/s

Answer 2

Explanation:

Average velocity is defined as the ratio of total displacement with respect to total time taken.

Given that, a particle moves along X-axis such that its position with respect to the origin is given  by x = bt², where b is 0.4 m/s².

We have to find the locity during the time interval t = 1 s to t = 5 s.

Average velocity = Total displacement/Total time

From above data displacement is x i.e. bt² and t is 1 to 5. Where 5s is final time & 1s is initial time.

So,

Average velocity = [b(5)² - b(1)²]/(5 - 1)

= (25b - b)/4

= 24b/4

= 6b

Also given that value of b is 0.4 m/s².

Therefore,

Average velocity = 6(0.4) = 2.4 m/s