Question

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 + k if (i) h = 0.1 (ii) h = 0.01 (iii) h = 0.001 Enter the exact answers.
(i) When h = 0.1, the average velocity between t = 1 and t = 1 + h is m/sec.
(ii) When h = 0.01, the average velocity between t = 1 and t = 1 + h is m/sec.
(iii) When h = 0.001, the average velocity between t = 1 and t = 1 + h is m/sec.
(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 instantaneous velocity appears to be m/sec.

Answer 1

(a)

i)
`V=12.6m/s`

ii)
`V=12.06m/s`

iii)
`V=12.006m/s`

(b)

`V = 10m/s`

Explanation:

Given

`s = 6t^2 + 4`

Solving (a): Average velocity between t = 1 and t = 1 + h

When t = 1

`t_1 = 1`

`s_1 = 6t^2 + 4 = 6 * 1^2 + 4 = 6 + 4 =10`

i) h = 0.1

When t = 1 + h

`t_2 = 1 + 0.1 = 1.1`

`s_2= 6t^2 + 4 = 6 * (1.1)^2 + 4 = 11.26`

Average velocity is then calculated as:

`V = (s_2 - s_1)/(t_2 - t_1)`

`V = (11.26 - 10)/(1.1- 1) = (1.26)/(0.1) = 12.6`

`V=12.6m/s`

ii) h = 0.01

When t = 1 + h

`t_2 = 1 + 0.01 = 1.01`

`s_2= 6t^2 + 4 = 6 * (1.01)^2 + 4 = 10.1206`

Average velocity is then calculated as:

`V = (s_2 - s_1)/(t_2 - t_1)`

`V = (10.1206 - 10)/(1.01- 1) = (0.1206)/(0.01) = 12.06`

`V=12.06m/s`

ii) h = 0.001

When t = 1 + h

`t_2 = 1 + 0.001 = 1.001`

`s_2= 6t^2 + 4 = 6 * (1.001)^2 + 4 = 10.012006`

Average velocity is then calculated as:

`V = (s_2 - s_1)/(t_2 - t_1)`

`V = (10.012006 - 10)/(1.001- 1) = (0.012006 )/(0.001) = 12.006`

`V=12.006m/s`

Solving (b): Instantaneous velocity at t = 1

When t = 1

`t = 1`

`s = 6t^2 + 4 = 6 * 1^2 + 4 = 6 + 4 =10`

Velocity is:

`V = (s)/(t)`

`V = (10)/(1)`

`V = 10m/s`