Respuesta :
Answer:
a) (i) When h=0.1, the average velocity between and is 12.6 m/sec.
(ii) When h=0.01, the average velocity between and is 12.06 m/sec.
(iii) When h=0.001, the average velocity between and is 12.006 m/sec.
b) The instantaneous velocity appears to be 12 m/sec.
Step-by-step explanation:
The question is incomplete:
In a time of t seconds, a particle moves a distance of S meters from its starting point, where S=6t^2+4.
(a) Find the average velocity between t=1 and t=1+h if: (i) h=0.1 (ii) h=0.01 (iii) h=0.001. Enter the exact answers.
(b) Use your answers to part (a) to estimate the instantaneous velocity of the particle at time t=1. Round your estimate to the nearest integer.
The average velocity can be expressed as:
[tex]\bar v=\frac{\Delta S}{\Delta t}=\frac{S(1+h)-S(1)}{(1+h)-1}= \frac{S(1+h)-S(1)}{h}[/tex]
For h=0.1:
[tex]\bar v= \frac{S(1+h)-S(1)}{h}=\frac{(6(1+h)^2+4)-(6(1)^2+4)}{h} = \frac{6(1^2+2h+h^2-1^2)+4-4}{h}\\\\\bar v = \frac{6h(2+h)}{h}=6(h+2)\\\\\\h=0.1\\\\\bar v=6(0.1+2)= 12.6[/tex]
For h=0.01
[tex]h=0.01\\\\\bar v=6(0.01+2)= 12.06[/tex]
For h=0.001
[tex]h=0.001\\\\\bar v=6(0.001+2)= 12.006[/tex]
b) As the h becomes close to zero, the average velocity tends to v=12 m/s
