Question

A body moving with uniform acceleration process distance of 65 metre in 5 secondsand 105 in 9th sec how far in 20 sec

Answer 1

$= > s_{nth} = 65 = u + a(n - \frac{1}{2})$

$= > 65 = u+a (\frac{9}{2})$-----1

Again,

$= > s_{nth} = 105 = u + a (n - \frac{1}{2})$

$= > 105 = u +a (\frac{17}{2})$

$= > 105 =u+ a (\frac{17}{2})$-----2

(2)-(1)

$105-65= u+a(\frac{17}{2})-u+a(\frac{9}{2})$

$=> 40 = (\frac{17a}{2}-\frac{9a}{2})$

$=> 40 = 4a$

$=> \frac{40}{4} = a$

$=> 10 = a$

Substituting it in eq---1

$=> 65 = u + a(n-\frac{1}{2})$

$=> 65 = u+ 10(5-\frac{1}{2}$

$=> 65 = u + 10 \times \frac{9}{2}$

$=> 65 = u + 5 \times 9$

$=> 65 - 45 = u$

$=> 20 = u$

$=> S = ut+ \frac{1}{2} a{t}^{2}$

$=> S = 20 x 20 + \frac{1}{2} \times 10 \times 20 \times 20$

$=> S = (400 + 2000) m$

$=> S = 2400 m$

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