Question

an athlete starts from rest with an acceleration 2 m/s ^2 and decelerates with 3 m/s ^2 if total length is 100 m then maximum speed attained by athlete is​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

The athlete have an acceleration ${ \alpha}$ = 2 m/s².

The athlete have an deceleration ${ \beta}$ = 3 m/s².

S = 100 m.

$\huge{\bold{\underline{Explanation:-}}}$

As you know,

${V_{max} = \frac{ \alpha \beta }{ \alpha + \beta } \times \sqrt{ \frac{2s( \alpha + \beta )}{ \alpha \beta } }}$

Simplifying will give.

$\huge{\boxed{\boxed{V_{max} = \sqrt{ \frac{2s \alpha \beta }{ \alpha + \beta } }}}}$

*very important formula.

Substituting the values.

${ V_{max} = \sqrt{ \frac{2 \times 100 \times 3 \times 2}{3 + 2} }}$

${ V_{max} = \sqrt{ \frac{2 \times 100 \times 3 \times 2}{5} }}$

${V_{max} = \sqrt{2 \times 20 \times 3 \times 2}}$

${V_{max} = \sqrt{6 \times 40}}$

${V_{max} = 4 \sqrt{30}}$

${ \therefore \sqrt{30} = 5.4}$

Substituting the values.

${ V_{max} = 4 \times 5.4}$

${V_{max} = 21.6 m/s}$

$\huge{\boxed{\boxed{V_{max} = 21.6 m/s}}}$

So,the maximum Speed reached by the athlete is 21.6 m/s.