A particle 'A' starts moving from point A with constant velocity 6m/s along x - axis. Another particle
'B' initially at rest starts moving along x- axis after (8/3) sec of 'A' from the same point, with an
acceleration a=
4m/s2
Based on the above information, answer the following questions.
The time when two particles will meet if they are moving in the same line
Answers
Explanation:
Given is constant velocity of A = 6m/s
acceleration of B = 4m/s and start moving after 8/3 sec of A.
so we have to find oit the position of A when b start moving, distance = time * speed
= 8/3*6
= 16m. along x axis.
______________________________
B.< 16m >A. x
let x is point where the A and B will meet in time t. and the distance between a and x is Y.
for B, s = ut + 1/2 at^2
16+y= 0*t+ 1/2*4*t^2
16+y = 0+ 2t^2
16+y = 2t^2————– (1)
for A t = d/v
t = y/ 6
6t = y ————————(2)
by putting value of y in (1) from (2)
16+6t = 2t^2
2 ( 8+3t ) = 2t^2
2/2(8+3t) = t^2
8 + 3t = t^2
t^2-3t-8 = 0
t = -b ±√b^2-4ac/2a [ formula]
t = -(-8) ± √(64-4*(-8)*1)/ 2
t = 8 ± √( 98)/2
t= (8± 9.59 )/2
so t = (8 + 9.59)/2 or (8-9.59)/2
t,= 8.795 second or -0.795 second
we know that time is always positive that's why the answer will be 8.795 second ≈ 8.8 second
i hope this answer is helpful to you!!!!
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