Question

Help me out to solve this problem because it looks difficult ​

Answer 1

Answer:

the time when he starts to deaccelerate will be =
`(8)/(5)`

distance between a and b at the starting = x-y = 12.6m

Step-by-step explanation:

writing equation of motion for traveller A v = u + at

where v is the final u the initial velocity and a is the acceleration

u = 10 v= 8 and a = -5 applying the values in the equation we get t =
`(2)/(5)`

therefore athlete A deaccelerates from 10 to 8 m/s in
`(2)/(5)`s

so the time when he starts to deaccelerate will be = 2 -
`(2)/(5)`

=
`(8)/(5)`

writing equation of motion for B

v = u + at

where v= 8 u = 0 and t = 2

therefore applying the values a= 4

`s = ut + (1)/(2) at^(2)`

therefore distance travelled be B in 2
`s_(1)` =
`s =x+   (1)/(2) 4×2^(2)`

therefore
`s_(1)` =x+ 8m

for traveller b distance travelled in 2
`s_(2)`= y + 10×
`(8)/(5)` + 10×
`(2)/(5)` +
`(1)/(2)`-5×
`((2)/(5) )^(2)`

= y+19.6 m

given
`s_(1) -s_(2) = 1`

therefore x-y-11.6 = 1

therefore distance between a and b at the starting = x-y = 12.6m