Question

A particle is moving with a uniform acceleration. Its velocity after 5 s is 25 ms and after 8 s is
4 ms. The distance travelled in the 12 second is
b) 55 m
a/50 m
c) 40.5 m
d) 44.5 m
​

Answer 1

EXPLANATION.

object is moving with uniform acceleration.

it's velocity after 5 seconds is 25 m

after 8 second is 34 m

To find distance travelled in the 12th

seconds.

$\sf : \implies \: from \: newton \:first \: equation \: of \: kinematics \\ \\ \sf : \implies \: v = u + at$

$\sf : \implies \: after \: five \: seconds \\ \\ \sf : \implies \: 25 = u + 5a \: \: \: ......(1) \\ \\ \sf : \implies \: after \: 8 \: seconds \\ \\ \sf : \implies \: 34 = u + 8a \: \: \: ......(2)$

$\sf : \implies \: from \: equation \: (1) \: \: \: and \: \: \: (2) \\ \\ \sf : \implies \: - 9 = - 3a \\ \\ \sf : \implies \: a = 3 \: ms {}^{ - 2} \\ \\ \sf : \implies \: put \: the \: value \: of \: (a )\: in \: \: equation \: (1)$

$\sf : \implies \: 25 = u + 5a \\ \\ \sf : \implies \: 25 = u + 5(3) \\ \\ \sf : \implies \: u \: = 10ms {}^{ - 1}$

$\sf : \implies \: formula \: of \: s_{n} \\ \\ \sf : \implies \: s_{n} = u + \frac{a}{2} (2n - 1) \\ \\ \sf : \implies \: distance \: travelled \: by \: object \: in \: 12th \: seconds\\ \\ \sf : \implies \: s_{n} = 10 + \frac{3}{2}(2 \times 12 - 1) \\ \\ \sf : \implies s_{n} = 10 + \frac{3}{2}(23) \\ \\ \sf : \implies \: s_{n} = 10 + 34.5 = 44.5 \: m \\ \\ \sf : \implies \: \green{{ \underline{distance \: travelled \: by \: object \: in \: 12th \: seconds \: = 44.5m}}}$