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nikdorinn [45]
3 years ago
5

In a time of t seconds, a particle moves a distance of s meters from its starting point, where s = 6t^2 + 4.

Mathematics
1 answer:
Lena [83]3 years ago
7 0

Answer:

(a)

i) V=12.6m/s

ii) V=12.06m/s

iii) V=12.006m/s

(b)

V = 10m/s

Step-by-step 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 = \frac{s_2 - s_1}{t_2 - t_1}

V = \frac{11.26 - 10}{1.1- 1} = \frac{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 = \frac{s_2 - s_1}{t_2 - t_1}

V = \frac{10.1206 - 10}{1.01- 1} = \frac{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 = \frac{s_2 - s_1}{t_2 - t_1}

V = \frac{10.012006 - 10}{1.001- 1} = \frac{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 = \frac{s}{t}

V = \frac{10}{1}

V = 10m/s

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