Question

A particle experiences a constant accelerating force after starting from rest. If it travels a distance x in the first two seconds and a distance y in next two seconds, then :

a) y = 4x
b) y = 3x
c) y = 2x
d) y = x

Answer 1

ok so it is y = x

according to "me" .. @mukushla

Answer 2

@UnkleRhaukus and @experimentX

Answer 3

y=x

Answer 4

That is totally wrong... I think ...

(:P sorry I did it wrong I think this now )

the body is having "accelerating " force... and hence y can't be equal to x

Answer 5

y > x

Answer 6

distance traveled in 2 seconds=x
distance traveled in 1 sec = x/2 similarly for y and when you'll equate this you'll have x=y ...time interval is equal and speed is not given so if we consider speed to be constant then y=x

Answer 7

but please answer my doubt that the body is having "accelerating" force and HENCE
We can not say that x = y ...

Answer 8

sorry didn't see that...so Y=2x

Answer 9

\[\large{h = ut + \frac{1}{2} at^2}\]
\[\large{x = \frac{1}{2}(a)(4)}\]
\[\large{x = 2a}\]

Answer 10

you forgot u

Answer 11

@ghazi u = 0 (given)

Answer 12

now similarly we have :

\[\large{h = ut+ \frac{1}{2}at^2}\]
\[\large{(x+y) = \frac{1}{2}(a)(16) }\]
\[\large{x+y = 8a}\]
\[\large{2a+y=8a}\]
\[\large{y = 6a}\]

Answer 13

x = 2a

y = 6a

x/y = 2a / 6a = 1/3

x = 3/y

y = 3x

Answer 14

is that completely right? or any mistake here

Answer 15

Though must say that idea came in mind when u suggested x/2 @ghazi

Answer 16

that seems fair :)

Answer 17

but for the next 2 seconds wouldn't there be some initial velocity

Answer 18

Note that I took x + y ... not y that's why u = 0