Answer :
Answer:
the time when he starts to deaccelerate will be =[tex]\frac{8}{5}[/tex]
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 = [tex]\frac{2}{5}[/tex]
therefore athlete A deaccelerates from 10 to 8 m/s in [tex]\frac{2}{5}[/tex]s
so the time when he starts to deaccelerate will be = 2 - [tex]\frac{2}{5}[/tex]
=[tex]\frac{8}{5}[/tex]
writing equation of motion for B
v = u + at
where v= 8 u = 0 and t = 2
therefore applying the values a= 4
[tex]s = ut + \frac{1}{2} at^{2}[/tex]
therefore distance travelled be B in 2 [tex]s_{1}[/tex] = [tex]s =x+ \frac{1}{2} 4×2^{2}[/tex]
therefore [tex]s_{1}[/tex] =x+ 8m
for traveller b distance travelled in 2 [tex]s_{2}[/tex]= y + 10×[tex]\frac{8}{5}[/tex] + 10×[tex]\frac{2}{5}[/tex] + [tex]\frac{1}{2}[/tex]-5×[tex](\frac{2}{5} )^{2}[/tex]
= y+19.6 m
given [tex]s_{1} -s_{2} = 1[/tex]
therefore x-y-11.6 = 1
therefore distance between a and b at the starting = x-y = 12.6m