Question

A body moves in a straight line along x axis its distance from the orgin is
given by the equation 8t -3t².The average velocity in the interval from t=0
to t=4 is​

Answer 1

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Answer 2

Answer:

$\underline{\boxed{v_{av} = \rm - 4 {ms}^{ - 1} }}$

Explanation:

In the given Question...

It is given that

x = 8t - 3t²

$v = \dfrac{dx}{dt}$

$v= \dfrac{d}{dt} \rm(8t - 3 {t}^{2} )$

$v= \rm \: 8 - 6t$

To find :-

The average velocity in the interval from t= 0 to t= 4s.

So,

When

t= 0

$v_{0} = \rm \: 8 - 6 \times 0$

$v_{0} = \rm \: 8 m {s}^{ - 1}$

At

t = 4s

$v_{4} = \rm \: 8 - 6 \times 4$

$v_{4} = \rm \: - 16 {ms}^{ - 1}$

The average velocity =

$v_{av} = \dfrac{ v_{i} +v_{f} }{2}$

$v_{av} = \dfrac{ v_{0} +v_{4} }{2}$

$v_{av} = \rm \dfrac{ 8 - 16 }{2}$

$v_{av} = \rm \dfrac{ - 8 }{2}$

$\underline{\boxed{v_{av} = \rm - 4 {ms}^{ - 1} }}$

Important Concepts--

$\leadsto \rm \: Average \: speed = \dfrac{ \Delta S}{\Delta t}$

$\leadsto \rm \: Average \: velocity= \dfrac{ \Delta \vec{S}}{\Delta t}$

$\leadsto \rm\: Average \: acceleration= \dfrac{ \Delta \vec{v}}{\Delta t}$