Answer:
Distance traveled during this acceleration will be 6950 m
Explanation:
Wear have given maximum speed tat will be equal to final speed of the car v = 278 m/sec
Constant acceleration 
As the car starts initially starts from rest so initial velocity of the car u = 0 m/sec
From third equation of motion 
Putting all values in equation
s = 6950 m
So distance traveled during this acceleration will be 6950 m
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
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Because it records speed of the car at a certain time, the independent variable should be time and dependent would be speed or velocity. Since it's taken every second, it would be considered instantaneous velocity, which is D.
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