Question

can any solve this the answer is b

Answer 1

Ans;_ 0.5m

Let  the particle  start  moving  from  point  A  and  due  to negative  acceleration it  stops moving  At  point  B.

From A to B

u = 9 m/s

v = 0

a = -2m/s²

t =?

Using  first  equation of motion,

v = u + at

0 = 9 - 2*t

t = 9/2

t = 4.5 second.

Distance  traveled in 4.5 second,

$S_{4.5}=ut+\frac{1}{2}at^{2}\\\\S_{4.5}=9*4.5-\frac{1}{2}2*(4.5)^{2}\\\\=40.5-20.25\\\\=20.25m$

Distance traveled in 4 second,

$S_{4}=ut+\frac{1}{2}at^{2}\\\\S_{4}=9*4-\frac{1}{2}2*(4)^{2}\\\\S_4=36-16\\\\S_4=20m$

Distance traveled C to B = 20.25 - 20 = 0.25m

Now,

After reaching  at  point  B, particle  will start  moving  in west  direction in direction of  acceleration.

We have  calculated  distance  traveled in 0.5 second  of 5th second. We have  to calculate remaining distance 0.5 m that  is  traveled in east  direction due  to acceleration.

For  motion from  east  to west,

u  = 0

a = +2 m/s²

t = 0.5

$S_{0.5}=ut+\frac{1}{2}at^{2}\\\\=0+\frac{1}{2}2.(0.5)^{2}\\ \\=0.25m$

Hence,

Total distance  traveled in 5th second  = 0.25 + 0.25 = 0.5 m