Question

a particle starts moving from rest state along a straight line under the action of a constant force and travels distance x in the first five seconds. The distance traveled by it in the next five seconds is

Answer 1

given u=0m/S
then
$using \: second \: eq \: of \: motion \\ s =ut + \frac{1}{2}a {t}^{2} \\ u = 0 \\ x = \frac{1}{2} \times a \times 5 \times 5 \\ 12.5a = x \\ for \: 10sec \: from \: start \\ let \: distance \: be \: y \: in \: 10sec \\ then \\ y = \frac{1}{2} \times a \times 10 \times 10 = 50a \\ \frac{x}{y} = \frac{12.5a}{50a} \\ 4x = y$
hence distance travelled in next 5 sec
= Total distance in 10sec-total distance in first five second
4x-x=3x
hence 3x is distance travelled