Question

Two particles, one with constant velocity of 50m/s and the other start from rest with uniform acceleration of 10m/s2.start moving simultaneously from the same position in the same direction. they will be at a distance of 125m from each other after? Option a) 5s and 5(1+√2)s. b)5s and 10s.

Answer 1

Relative motion concepts will be applied here.

constant velocity of first particle u = 50ms-1

Acceleration a = 10 ms-2

Distance Srel = s + ut + 1/2 at^2

S rel = 125 + 50t + 1/210t^2

Srel will be 0

0 = 125+(-50t) + 5t^2

now devide equation by 5

0 = 25- 10t + t^2

we can write it as

t^2 -10t + 25 = 0

(t-5)^2 = 0

take roots both the sides

then
t - 5 = 0

t = 5s is our answer


Hope it's useful

Answer 2

use relative concept of kinematic
Srel= Sorel +urel.t + 1/2 arel.t^2
sorel is initial position of A and B
where Srel is final positin of A w.r.t. B
urel velocity of A w.r.t B
and arel acceleration of A w.r.t B
when they cross then Srel=0
0= 125+ (50-0) t+1/2 (0-10) t^2
0=25 +10t-t^2
t^2-10t-25=0
t={10+_root (200)}/2
=5+_5root2
hence t =5(1+root2) sec