Diagram 7 shows the velocity-time graph of an object for 12 s.

When is the object at a distance of one-quarter of
the total distance travelled in the 12 s?
A 3 s
B 4 s
C 5 s
D 6 s

*The answer is D but can anyone tell me the solving steps? Or perhaps the answer is wrong?*

t = 6

Explanation:

Displacement is equal to the area under a velocity vs time graph.

In this case, the area is a triangle.  At time t, the base of the triangle is t.  The height of the triangle can be found with similar triangles:

h / t = 8 / 12

h = ⅔ t

So the distance traveled at time t is:

d = ½ (t) (⅔ t)

d = ⅓ t²

The distance traveled at time 12 is:

D = ½ (12) (8)

D = 48

We want to find when d = D/4.

d = D/4

⅓ t² = 48/4

⅓ t² = 12

t² = 36

t = 6

Alternatively, since the acceleration is constant here, we could use a constant acceleration equation.

Δx = v₀ t + ½ at²

Given v₀ = 0 m/s and a = ⅔ m/s²:

Δx = (0) t + ½ (⅔) t²

Δx = ⅓ t²

When t = 12, Δx = 48.

⅓ t² = 48/4

t = 6

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!