Question

A particle moving with uniform retardation covers distances 18m, 14m, and 10m in successive seconds. it comes to rest after travelling a further distance of how many metres?

Answer 1

Quick method:

    Let the initial velocity 0 of an object.  Let that accelerate:   Then  x = 1/2 a t^2.

We see that distances travelled in ‘t’ th second:   (t-1/2) a

s_1 = a/2            s_2 = 3a/2       s_3= 5a/2       s_4= 7a/2

They are in ratio:   1 : 3 : 5 : 7 : 9 : 11 : 13  ... where constant of ratio = a/2.

Given distances are: 18 : 14 : 10   ie.,   in the ratio  9 : 7 : 5

So the object will travel two more seconds and covers a distance of  1d+3d before stopping, where 9d = 18m.  ie., another 8 meters.

Note: we cannot apply this method, if the distances given are not in the above ratios.

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Long method:

Let  us say that the distance of 18 m is covered during n’th second. During t = n-1 to n sec.    Let the velocity at t =0  be u  and deceleration be a.

$x = u t - 1/2 a t^2\\x_{n-1} = un - u - a (n-1)^2/2 = un - u - an^2/2 + na - 1/2 a\\\\x_n = u n - a n^2/2\\\\x_n+1 = un +u - a (n+1)^2/2 = un + u - an^2/2 - n a - a/2\\\\s_n = x_n - x_n+1 = u - (n - 1/2) a = 18 m ---(1)\\s_n+1 = x_n+1 - x_n = u - (n+ 1/2 )a = 14 m --- (2)\\\\(1) -(2) =\ \textgreater \ a = 4 m/s^2\\(1) =\ \textgreater \ u = 16 + 4 n = 4(n+4)\\\\Speed\ at\ t = n+2\ is\ u - 4 (n+2) = 8 m/s$

So final velocity will be 0 in (8-0)/4 = 2 sec.

Distance travelled after t= n+2 sec,  in 2 sec.

              s = 8 *2 - 1/2 4 *2^2 = 8 meters

Answer 2

Answer:

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