Me out to solve this problem because it looks difficult

the time when he starts to deaccelerate will be =

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 =

therefore athlete A deaccelerates from 10 to 8 m/s in s

so the time when he starts to deaccelerate will be = 2 -

                                                                                    =

writing equation of motion for B

v = u + at

where v= 8 u = 0 and t = 2

therefore applying the values a= 4

therefore distance travelled be B in 2 =

therefore   =x+ 8m

for traveller b distance travelled in 2  = y + 10× + 10× + -5×

= y+19.6 m

given

therefore x-y-11.6 = 1

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

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles