Answer:
A) The acceleration is zero
<em>B) The total distance is 112 m</em>
Explanation:
<u>Velocity vs Time Graph</u>
It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.
The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.
A. To calculate the acceleration, we use the formula:

Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:

This confirms the previous conclusion.
B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.
Area of rectangle= base*height=12 s * 8 m/s = 96 m
Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m
The total distance is: 96 m + 16 m = 112 m
Complete Question
Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?
Answer:
Go-cart A is faster
Explanation:
From the question we are told that
The length of the track is 
The speed of A is 
The uniform acceleration of B is 
Generally the time taken by go-cart A is mathematically represented as
=> 
=> 
Generally from kinematic equation we can evaluate the time taken by go-cart B as

given that go-cart B starts from rest u = 0 m/s
So

=>
=>
Comparing
we see that
is smaller so go-cart A is faster
Answer:
rotates faster
Explanation:
A huge rotating cloud of particles in space gravitate together to form an increasingly dense ball As it shrinks in size, the cloud rotates faster. Because Angular momentum is conserved, so when it shrinks the moment of inertia decreases, then angular speed must increase. So it rotates fast.