Question

A particle is moving with uniform acceleration covers a distance of $$45$$ m in $$6^{th}$$ second and $$75$$ m in $$12^{th}$$ second during its motion. Calculate the displacement of the particle after the $$20s$$?

Answer 1

$$S_6 = 45m, S_{12}=75m$$
using $$S_n = u + a (n – 1/2 )$$
i.e., $$S_6 = u + a$$ or $$45 = u + 5.5a$$ ……… (1)
$$S_{12}= u + a(12 – 1/2)$$ or $$75 = u + a(11.5)$$ …….. (2)
Eq(2) – (1) gives $$6 a = 30$$
Or $$a = 30/6 = 5m/s^2$$
From (1), $$45 = u + 5 \times 5.5$$
$$u = 17.5m/s.$$
Distance travelled in $$t = 20s$$,
$$S = ut+ 1/2at^2$$
$$= 17.5 \times 20 + 1/2 \times 5 \times 20^2$$
$$=1350m.$$