Question

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

Answer 1

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:

$V = \frac{s}{t}$

$V = \frac{10}{1}$

$V = 10m/s$