Automobile a starts from o and accelerates at the constant rate of 0.75 m/s2. a short time later it is passed by bus b which is traveling in the opposite direction at a constant speed of 6 m/s. knowing that bus b passes point o 20 s after automobile a started from there, determine when and where the vehicles passed each other.

they meet from point o at distance 50.46 m and time taken is 11.6 seconds

Explanation:

given data

acceleration = 0.75 m/s²

speed B = 6 m/s

time B = 20 s

to find out

when and where the vehicles passed each other

solution

we consider here distance = x , when they meet after o point

and time = t for meet point z

we find first Bus B distance for 20 s ec

distance B = velocity × time

distance B = 6 × 20

distance B = 120 m

so

B take time to meet is calculate by distance formula

distance = velocity × time

120 - x = 6 × t

x = 120 - 6t   1

and

distance of A when they meet by distance formula

distance = ut + 1/2 × at²

here u is initial speed = 0 and t is time

x = 0 + 1/2 × 0.75 × t²

x = 0.375 × t²      2

so from equation 1 and 2

0.375 × t²  = 120 - 6t

t = 11.6

so time is 11.6 second

and

distance from point o from equation 2

x = 0.375 (11.6)²

x = 50.46

so distance from point o is 50.46 m

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.

i think the answer is tomato juice? not entirely sure though