The velocity-time graph of a particle in linear motion is shown in figure. Both v and t are in SI units. What is the displacement of the particle from the origin after 8 second?

Question

The velocity-time graph of a particle in linear motion is shown in figure. Both v and t are in SI units. What is the displacement of the particle from the origin after 8 second? (A) 6 m (B) 8 m (C) 16 m (D) 18 m

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/30f527ab1260b7f12c1592bf2d063240-.png" /> Displacement in 8 s =+ Area O A B C- Area C D E F =(1/2)(A B+O C) × altitude -(1/2)(C F+D E) × altitude =(1/2)(2+4) ⋅(4)-(1/2)(4+2) × 2=12-6=6 m

Q. The velocity-time graph of a particle in linear motion is shown in figure. Both $v$ and $t$ are in SI units. What is the displacement of the particle from the origin after 8 second?

6 m

8 m

16 m

18 m

Displacement in $8 s = +$ Area $O A BC -$ Area $C D EF$ $= \frac{1}{2} (A B + OC) \times$ altitude $- \frac{1}{2} (CF + D E) \times$ altitude
$= \frac{1}{2} (2 + 4) \cdot (4) - \frac{1}{2} (4 + 2) \times 2 = 12 - 6 = 6 m$

Q. The velocity-time graph of a particle in linear motion is shown in figure. Both v and t are in SI units. What is the displacement of the particle from the origin after 8 second?

6 m

8 m

16 m

18 m

Displacement in 8s=+ Area OABC− Area CDEF =21​(AB+OC)× altitude −21​(CF+DE)× altitude =21​(2+4)⋅(4)−21​(4+2)×2=12−6=6m

Q. The velocity-time graph of a particle in linear motion is shown in figure. Both $v$ and $t$ are in SI units. What is the displacement of the particle from the origin after 8 second?

Displacement in $8 s = +$ Area $O A BC -$ Area $C D EF$ $= \frac{1}{2} (A B + OC) \times$ altitude $- \frac{1}{2} (CF + D E) \times$ altitude
$= \frac{1}{2} (2 + 4) \cdot (4) - \frac{1}{2} (4 + 2) \times 2 = 12 - 6 = 6 m$