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

Help me out to solve this problem because it looks difficult ​-example-1
User Scomes
by
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1 Answer

6 votes

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

User Piyush Kashyap
by
8.3k points

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