$\"\"$

The velocity of the rock is thrown upward on the planet is $\"10\"$ m/s.

The function of the ball is $\"y=10t-1.86t^2\"$ .

(a)

Average velocity :

The average velocity is defined as the ratio of change in distance by change in time.

$\"v_{avg}=\\frac{\\Delta$ , where $\"\\Delta$ and $\"\\Delta$ .

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(i)

Find the average velocity over the time interval $\"[1,$ .

$\"v_{avg}=\\frac{\\Delta$

$\"v_{avg}=\\frac{y(t_2)-y(t_1)}{t_2-t_1}\"$

Substitute $\"t_1=1\"$ and $\"t_2=2\"$ in the above expression.

$\"\\\\v_{avg}=\\frac{y(2)-y(1)}{2-1}\\\\\"$

$\"\\\\=\\frac{[10(2)-1.86(2)^2]-[10(1)-1.86(1)^2]}{2-1}\\\\$

$\"\\\\=4.42\"$

The average velocity over the time interval $\"[1,$ is $\"4.42\"$ m/s.

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(ii)

Find the average velocity over the time interval $\"[1,$ .

$\"v_{avg}=\\frac{\\Delta$

$\"v_{avg}=\\frac{y(t_2)-y(t_1)}{t_2-t_1}\"$

Substitute $\"t_1=1\"$ and $\"t_2=1.5\"$ in the above expression.

$\"\\\\v_{avg}=\\frac{y(1.5)-y(1)}{1.5-1}\\\\\"$

$\"\\\\=\\frac{[10(1.5)-1.86(1.5)^2]-[10(1)-1.86(1)^2]}{1.5-1}\\\\$

$\"\\\\=5.35\"$

The average velocity over the time interval $\"[1,$ is $\"5.35\"$ m/s.

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(iii)

Find the average velocity over the time interval $\"[1,$ .

$\"v_{avg}=\\frac{\\Delta$

$\"v_{avg}=\\frac{y(t_2)-y(t_1)}{t_2-t_1}\"$

Substitute $\"t_1=1\"$ and $\"t_2=1.1\"$ in the above expression.

$\"\\\\v_{avg}=\\frac{y(1.1)-y(1)}{1.1-1}\\\\\"$

$\"\\\\=\\frac{[10(1.1)-1.86(1.1)^2]-[10(1)-1.86(1)^2]}{1.1-1}\\\\$

$\"\\\\=6.094\"$

The average velocity over the time interval $\"[1,$ is $\"6.094\"$ m/s.

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(iv)

Find the average velocity over the time interval $\"[1,$ .

$\"v_{avg}=\\frac{\\Delta$

$\"v_{avg}=\\frac{y(t_2)-y(t_1)}{t_2-t_1}\"$

Substitute $\"t_1=1\"$ and $\"t_2=1.01\"$ in the above expression.

$\"\\\\v_{avg}=\\frac{y(1.01)-y(1)}{1.01-1}\\\\\"$

$\"\\\\=\\frac{[10(1.01)-1.86(1.01)^2]-[10(1)-1.86(1)^2]}{1.01-1}\\\\$

$\"\\\\=6.2614\"$

The average velocity over the time interval $\"[1,$ is $\"6.2614\"$ m/s.

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(v)

Find the average velocity over the time interval $\"[1,$ .

$\"v_{avg}=\\frac{\\Delta$

$\"v_{avg}=\\frac{y(t_2)-y(t_1)}{t_2-t_1}\"$

Substitute $\"t_1=1\"$ and $\"t_2=1.001\"$ in the above expression.

$\"\\\\v_{avg}=\\frac{y(1.001)-y(1)}{1.001-1}\\\\\"$

$\"\\\\=\\frac{[10(1.001)-1.86(1.001)^2]-[10(1)-1.86(1)^2]}{1.001-1}\\\\$

$\"\\\\=6.2784\"$

The average velocity over the time interval $\"[1,$ is $\"6.2784\"$ m/s.

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(b)

Find the instantaneous velocity when $\"t=1\"$ .

Instantaneous velocity :

The instantaneous velocity is defined as $\"v=\\lim_{\\Delta$ .

$\"v=\\lim_{\\Delta$

Substitute $\"t=1\"$ in the above expression.

$\"v=\\lim_{\\Delta$

$\"=\\lim_{\\Delta$ ( Since $\"y(t)=10t-1.86t^2\"$ )

$\"=\\lim_{\\Delta$

$\"\\small$

$\"=\\lim_{\\Delta$

$\"=\\lim_{\\Delta$

$\"=\\lim_{\\Delta$

$\"\\\\={6.28+1.86(0)}\\\\$

The instantaneous velocity when $\"t=1\"$ is $\"6.28\"$ m/s.

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(a) (i) The average velocity over the time interval $\"[1,$ is $\"4.42\"$ m/s.

(ii) The average velocity over the time interval $\"[1,$ is $\"5.35\"$ m/s.

(iii) The average velocity over the time interval $\"[1,$ is $\"6.094\"$ m/s.

(iv) The average velocity over the time interval $\"[1,$ is $\"6.2614\"$ m/s.

(v) The average velocity over the time interval $\"[1,$ is $\"6.2784\"$ m/s.

(b) The instantaneous velocity when $\"t=1\"$ is $\"6.28\"$ m/s.