Two objects, A and B, accelerate from rest at uniform rates. B accelerates twice as much as Compared to A, B travels

Toptutor

2 years ago

uA=0 uB=0 ( since they both started from rest)
aA= a aB=2a( acceleration of B is twice that of A)
From S=ut+1/2at^2
SA= 0+1/2atA^2
sA=1/2at^2 ( distance covered by body A)
Also, SB=1/2*2atB^2( 2 cancels out 2)
sB= at^2
Since the two bodies started at uniform rate(tA=tB)
making tA the subject of formula from SA above
tA=√2SA/a
Also tB= √SB/a
Equate tA and tB
√2sA/a= √SB/a ( cancel out the square roots and a)
we have: 2SA=SB
Therefore B travels two times as far