Question

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

Answer 1

Answer:

the time when he starts to deaccelerate will be =$\frac{8}{5}$

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

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 = $\frac{2}{5}$

therefore athlete A deaccelerates from 10 to 8 m/s in $\frac{2}{5}$s

so the time when he starts to deaccelerate will be = 2 - $\frac{2}{5}$

                                                                                    =$\frac{8}{5}$

writing equation of motion for B

v = u + at

where v= 8 u = 0 and t = 2

therefore applying the values a= 4

$s = ut + \frac{1}{2} at^{2}$

therefore distance travelled be B in 2 $s_{1}$ = $s =x+ \frac{1}{2} 4×2^{2}$

therefore  $s_{1}$ =x+ 8m

for traveller b distance travelled in 2  $s_{2}$= y + 10×$\frac{8}{5}$ + 10×$\frac{2}{5}$ + $\frac{1}{2}$-5×$(\frac{2}{5} )^{2}$

= 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