Question

A particle starts from rest and moves with uniform acceleration. Show that distance travelled in 1st second, 2nd

second and 3rd second are in the ratio of 1 : 3 : 5.​

Answer 1

This is known as Galileo's law of odd numbers. For every second, the distance ratio would be 1 : 3 : 5 : 7 : 9... and so on.

Answer:

Distance covered in 'nth' second with uniform acceleration 'a' and initial velocity 'u' is:

$S_{nth} = u + \dfrac{a}{2}(2n - 1)$

For 1st second:

S1st = 0 + a/2(2 - 1) = a/2 __(1)

For 2nd second:

S2nd = 0 + a/2(2 × 2 - 1) = 3a/2 ___(2)

For 3rd second:

S3rd = 0 + a/2(2 × 3 - 1) = 5a/2 ____(3)

Now, ratio of RHS of eqn. (1), (2), & (3):-

a/2 : 3a/2 : 5a/2

= 1 : 3 : 5

H.P.