Question #de7ca

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Answer 1 · 2016-12-01T05:38:25

We know that the distance traveled by the body in *n th sec is given by the equation

$S_{n} = u + \frac{1}{2} a (2 n - 1)$ $S_n=u+1/2a(2n-1)$

where u is the initial velocity and a is the uniform acceleration of the body.

In our given problem $u = 0$ $u=0$

So distance traveled in 3rd sec

$S_{3} = \frac{1}{2} a (2 \cdot 3 - 1) = \frac{5}{2} a$ $S_3=1/2a(2*3-1)=5/2a$

And distance traveled in 4th sec

$S_{4} = \frac{1}{2} a (2 \cdot 4 - 1) = \frac{7}{2} a$ $S_4=1/2a(2*4-1)=7/2a$

So the ratio of distance traveled is

$\frac{S_{3}}{S_{4}} = \frac{5}{7}$ $S_3/S_4=5/7$