Answered

IDNLearn.com makes it easy to find precise answers to your specific questions. Discover in-depth and trustworthy answers to all your questions from our experienced community members.

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.

Sagot :

MrRoyalidn

Answer:

(a)

i) [tex]V=12.6m/s[/tex]

ii) [tex]V=12.06m/s[/tex]

iii) [tex]V=12.006m/s[/tex]

(b)

[tex]V = 10m/s[/tex]

Step-by-step explanation:

Given

[tex]s = 6t^2 + 4[/tex]

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

When t = 1

[tex]t_1 = 1[/tex]

[tex]s_1 = 6t^2 + 4 = 6 * 1^2 + 4 = 6 + 4 =10[/tex]

i) h = 0.1

When t = 1 + h

[tex]t_2 = 1 + 0.1 = 1.1[/tex]

[tex]s_2= 6t^2 + 4 = 6 * (1.1)^2 + 4 = 11.26[/tex]

Average velocity is then calculated as:

[tex]V = \frac{s_2 - s_1}{t_2 - t_1}[/tex]

[tex]V = \frac{11.26 - 10}{1.1- 1} = \frac{1.26}{0.1} = 12.6[/tex]

[tex]V=12.6m/s[/tex]

ii) h = 0.01

When t = 1 + h

[tex]t_2 = 1 + 0.01 = 1.01[/tex]

[tex]s_2= 6t^2 + 4 = 6 * (1.01)^2 + 4 = 10.1206[/tex]

Average velocity is then calculated as:

[tex]V = \frac{s_2 - s_1}{t_2 - t_1}[/tex]

[tex]V = \frac{10.1206 - 10}{1.01- 1} = \frac{0.1206}{0.01} = 12.06[/tex]

[tex]V=12.06m/s[/tex]

ii) h = 0.001

When t = 1 + h

[tex]t_2 = 1 + 0.001 = 1.001[/tex]

[tex]s_2= 6t^2 + 4 = 6 * (1.001)^2 + 4 = 10.012006[/tex]

Average velocity is then calculated as:

[tex]V = \frac{s_2 - s_1}{t_2 - t_1}[/tex]

[tex]V = \frac{10.012006 - 10}{1.001- 1} = \frac{0.012006 }{0.001} = 12.006[/tex]

[tex]V=12.006m/s[/tex]

Solving (b): Instantaneous velocity at t = 1

When t = 1

[tex]t = 1[/tex]

[tex]s = 6t^2 + 4 = 6 * 1^2 + 4 = 6 + 4 =10[/tex]

Velocity is:

[tex]V = \frac{s}{t}[/tex]

[tex]V = \frac{10}{1}[/tex]

[tex]V = 10m/s[/tex]

We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.