In a time of t seconds, a particle moves a distance of...

Question

In a time of t seconds, a particle moves a distance of s meters from its starting point, where s equals 6t squared plus 4.

(a) Find the average velocity between t equals 1 and t equals 1 plus h if
(i) h equals 0.1 (ii) h equals 0.01 (iii) h equals 0.001

Answer 1

Average velocity = Δs/Δt

 

This seems to be simply a plug-in.

 

Δs = s(1 + h) - s(1)

Δt = t_f - t_i = (1 + h) - 1 = h

 

where s = 6t² + 4.

 

For part (i), h = 0.1, so

 

s(1) = 10, s(1 + h) = 6*1.1² + 4 = 11.26, and so

Δs = 11.26 - 10 = 1.26 

 

avg v = 1.26/0.1 = 12.6 m/s

 

Parts (ii) and (iii) follow the same pattern