A body moving with a uniform acceleration crosses a distance of $65 \mathrm{~m}$ in the 5 th second and $105 \mathrm{~m}$ in 9 th second. How far will it go in 20 s?

Question

Aicharya · Accepted Answer

We have, $S_{n}=u+\frac{a}{2}(2 n-1)$
or $65=\mathrm{u}+\frac{a}{2}(2 \times 5-1)$
or $65=\mathrm{u}+\frac{9}{2} a$
Also, $105=\mathrm{u}+\frac{a}{2}(2 \times 9-1)$
or $105=\mathrm{u}+\frac{17}{2} a$
Equation $(2)-(1)$ gives,
$40=\frac{17}{2} a-\frac{9}{2} a=4 a$ or $a=10 \mathrm{~m} / \mathrm{s}^{2}$
Substitute this value in (1) we get, $\mathrm{u}=65-\frac{9}{2} \times 10=65-45=20 \mathrm{~m} / \mathrm{s}$
$\therefore$ The distance travelled by the body in $20 \mathrm{~s}$ is,
$\mathrm{s}=\mathrm{ut}+\frac{1}{2} a \mathrm{t}^{2}=20 \times 20+\frac{1}{2} \times 10 \times(20)^{2}$
$=400+2000=2400 \mathrm{~m}$

A body moving with a uniform acceleration crosses a distance of $65 \mathrm{~m}$ in the 5 th second and $105…

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