Question

A body, moving in a stragiht line with an initial velocity of 5 ms-' and a constant acceleration, covers a
distance of 30 m in the 3rd second. How much distance will it cover in the next 2 second
(A)
70 m
(B)
80 m
(C)
90 m
(D)
100m

# 90 is answer but how explain it plsss...​

Answer 1

Answer:

formula used

$s(t) = u + \frac{1}{2} a(2t - 1)$

now .....

$s(3) = 5 + \frac{1}{2} a(2 \times 3 - 1) \\ 30 = 5 + \frac{5a}{2} \\ 60 = 10 + 5a \\ a = 10m.s {}^{ - 1}$

now distance covered in the next two seconds is

$s(5) = 5 + \frac{1}{2} \times 10(2 \times 5 - 1) \\ = > s(5) = 5 + 45 = 50 \: \: m$

$s(4) = 5 + \frac{1}{2} \times 10(2 \times 4 - 1) = 5 + 35 = 40 \: m$

total distance in next two seconds is

=s(4)+s(5)=(40+50)=90 metres