The variation of velocity of a particle moving along a straight line is shown in figure. The distance traversed by the particle in 4 second is

Question

The variation of velocity of a particle moving along a straight line is shown in figure. The distance traversed by the particle in 4 second is (A) 60 m (B) 25 m (C) 55 m (D) 30 m

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f764097ff866213e7b0717018a00d158-.png" /> Distance travelled in 4 s = Area under the curve = area of Δ A O H+ area of rectangle A B G H+ area of Δ B C C' + area of rectangle C' D E G =(20 × 1/2)+20 × 1+(10 × 1/2)+10 × 2=55 m

Q. The variation of velocity of a particle moving along a straight line is shown in figure. The distance traversed by the particle in 4 second is

60 m

25 m

55 m

30 m

Distance travelled in $4 s =$ Area under the curve
$=$ area of $Δ A O H +$ area of rectangle $A BG H +$ area of $Δ BC C^{'}$
$+$ area of rectangle $C^{'} D EG$
$= \frac{20 \times 1}{2} + 20 \times 1 + \frac{10 \times 1}{2} + 10 \times 2 = 55 m$

Q. The variation of velocity of a particle moving along a straight line is shown in figure. The distance traversed by the particle in 4 second is

60 m

25 m

55 m

30 m

Distance travelled in 4s= Area under the curve = area of ΔAOH+ area of rectangle ABGH+ area of ΔBCC′ + area of rectangle C′DEG =220×1​+20×1+210×1​+10×2=55m

Distance travelled in $4 s =$ Area under the curve
$=$ area of $Δ A O H +$ area of rectangle $A BG H +$ area of $Δ BC C^{'}$
$+$ area of rectangle $C^{'} D EG$
$= \frac{20 \times 1}{2} + 20 \times 1 + \frac{10 \times 1}{2} + 10 \times 2 = 55 m$