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

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