Question

a particle position as a function of time is given as x= 4t2+6t+8 find average.velocity of particle t=0 to t= 6s​

Answer 1

Answer:

14

Explanation:

4*0*2=0+6+8

=6+8=14

Answer 2

Explanation:

$x = 4 {t}^{2} + 6t + 8$

As we know velocity is give by

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

therefore

$v = \frac{d}{ dt} (4 {t}^{2} + 6t + 8)$

$v = 8t + 6$

velocity at t=0s is

$v = 8\times 0 + 6$

velocity

$v = 6 \frac{m}{s}$

at t=6s is

$v = 8 \times 6 + 6$

$v = 54\frac{m}{s}$

therefore average velocity is

V_av=total distance/total time

V_av=(54-6)/(6-0)

V_av=8m/s