Answer:
I think It Should be Star B
Explanation:
We have: v i (initial velocity) = 6 m/sv = 1.1 m/sa = - 9.8 m/s²v = v i + a · t1.1 m/s = 6 m/s - 9.8 m/s² t9.8 t = 6 - 1.19.8 t = 4.9t = 4.9 : 9.8t = 0.5 sThen the replacement:x = xi + vi · t + a t² / 2( xi = 0 )x = 6 · 0.5 - 9.8 · 0.25 / 2x = 3 - 1.225Answer:
x = 1.775 m
As per kinematics equation we know that
final speed of the car = 0 m/s
initial speed is given as 30 m/s
distance moved = 100 m
now we have



now braking force is given as

now for mass we know that the weight of car is

so mass of car is

now we have

Part b)
Again we have
final speed of the car = 0 m/s
initial speed is given as 30 m/s
distance moved = 10 m
now we have



now braking force is given as

mass of car is

now we have

Total buns = 21 1/2 = 43/2 buns
1/3 size buns would be: 43/2 / 1/3 = 43*3 / 2 = 129/2
In short, Your Answer would be 129/2 or 64 1/2 or 64.5
Hope this helps!
Use the kinematic equation,
Vf^2 = Vi^2 + 2aX
31.5^2 = 11.6^2 + 2(5.22)(x)
Solve for x
Answer: 82.2m