Question

two particles A and B are moving in the same direction on same straight line a is ahead of B by 20m and A has constant speed 5M per second and B has initial speed 30 M per second and retardation of 10 M per second square then if x is a total distance travelled by B as it meets a for the second time then value of X is

Answer 1

velocity of b with respect to a=30-5=25m/s
acceleration of b w.r.t.a=-10-0=-10m/s^2
s=20m=ut+1/2*a*t^2=25t-5t^2
t^2-5t+4=0
then t=4sec &t=1sec
distance travelled by b=x=ut+1/2*a*t^2
=30t-5t^2
at t=1 sec
x=30-5=25m
qt t=4sec
x=120-80=40m
so 40 m is answer.

Answer 2

Answer:

15 m

Explanation:

Two particles A and B are moving in the same direction on same straight line.

A is ahead of B by 20 m.

For A,

Speed = 5 m/s

Let in t seconds A meets B two times.

Time = t

Distance covered by A, d = 5t

For B,

Speed  = 30 m/s

Retardation = 10 m/s²

Time = t

$s=ut-\dfrac{1}{2}at^2$

$s=30t-5t^2$

Distance covered by B, s

A is ahead of B by 20 m.

Both will meet, when d+20 = s    (A is ahead of B by 20 m)

$5t+20=30t-5t^2$

$5t^2-25t+20=0$

$t^2-5t+4=0$

$t=1,5$

A and B meet at t=1 and t=4

Distance covered by A in t=1 sec, 5 m

Distance covered by A in t=4 sec, 20 m

x represents total distance traveled by B as it meets A for second time.

x = 20 - 5

x = 15 m

Hence, They meet two times between 15 m