Instantaneous rate of change

Question

In a time of t seconds, a particle moves a distance of s meters from its starting point, where s equals 4t squared plus 2.

(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

 

Use your answers to part (a) to estimate the instantaneous velocity of the particle at time t equals 1.

Round your estimate to the nearest integer.

Answer 1

Hi Brooklyn,

 

s(t) = 4t² + 2

(a) The average velocity between t = 1 and t = 1 +h is

 

V(average)= (s(1+h) - s(1) )/(h +1 - 1),

s(1+h) = 4(1 + h)² + 2 = 4(1+2h+h²)+2 = 6+8h+4h²

s(1) = 4(1)² + 2= 6

 

V(average)=(6+8h+4h² -6)/h = 8+4h meter/second

(i) h = 0.1

v = 8 + 4h = 8+4*0.1= 8.4 meters/second ~ 8 meter/second

(ii) h = 0.01

v = 8 + 4h = 8+4*0.01 = 8.04 meter/second ~ 8 meter/second

(iii) h =0.001

v = 8 + 4h = 8 + 4*0.001 = 8.004 meter/second ~ 8 meter/second