Question

A particle is uniformly accelerated from A to B a distance of 192 m and is then uniformly retarded from B to C a distance of 60 m the speeds of particle at A and B are 4m/s and V m/s total time 22 s value of V ?

Answer 1

A=============B===========C
let a is the acceleration of particle .
t is the time taken by particle.
v=u + at
V= 4 +at

also ,
v^2 = u^2 + 2aS
V^2 =4^2 +2.a.192
V^2 = 16 + 384 a
put V= 4 + at
(4 + at )^2 = 16 + 384 a
16 + 8at + at^2 = 16 + 384 a

at^2 +8at -384a =0
t^2 + 8t -384 =0
t = {-8+_root (64 +4 x 384)/2
=( -8 +_40)/2 =16 , -24

t=-24 isn't possible
hence t = 16 sec
it means particle covered AB distance take time 16sec

for B to C
let b is the retardation of particle then ,
and particle 22-16 =6 sec in motion .
use formula ,
v =u +bt
finally velocity of particle =0
0 =V -6b
V=6b
again ,
v^2 =u^2 +2bS"
0 = V^2 -2b.60
V^2 =120b
put V=6b

36b^2 =120b
b= 120/36 =10/3 m/sec^2
now
V=6b =6 x 10/3 = 20 m/sec

Answer 2

A - B:    u = 4 m/s    v = V m/s       s = 192 m
             acceleration  a = (v² - u²)/2s = (V² - 16)/384   m/s²
             t1 = (v-u)/a = 384/(V+4)  sec

B - C:   u = V m/s   s = 60 m     v = 0
            a = (v² - u²)/2s = - V²/120
            t2 = (v-u)/a = s / [ (v+u)/2 ] = 120/V  sec

Given  t1 + t2 = 22 sec
           384/(V+4)  + 120/V = 22 sec
           11 V (V+4) = 192 V + 60 (V+4)
           11 V² - 208 V - 240 = 0
            V = [ 104 +- √(104² + 2640) ] / 11
               = 20 m/s  ignoring -ve value.