The acceleration vs distance graph for a particle moving with initial velocity 5 m / s is shown in the figure. The velocity of the particle at x=35 m will be

Question

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/259a6b7373fe32bb9f4f88232133a31c-.png" /> The acceleration vs distance graph for a particle moving with initial velocity 5 m / s is shown in the figure. The velocity of the particle at x=35 m will be (A) 20.62 m/s (B) 20 m/s (C) 25 m/s (D) 50 m/s

Tardigrade Admin · Accepted Answer

Area of graph =∫ a d x=∫ (v d v/d x) ⋅ d x==∫Vi v d v =[(v2/2)]vivt, 150+150=(1/2)[Vf2-Vi2] 300 × 2=Vf2-(5)2 Vf2=600+25, Vf=25

Q.
The acceleration vs distance graph for a particle moving with initial velocity $5 m / s$ is shown in the figure. The velocity of the particle at $x = 35 m$ will be

20.62 m/s

20 m/s

25 m/s

50 m/s

Area of graph
$= \int a d x = \int \frac{v d v}{d x} \cdot d x == \int_{V_{i}} v d v$
$= [\frac{v ^{2}}{2}]_{v_{i}}^{v_{t}}, 150 + 150 = \frac{1}{2} [V_{f}^{2} - V_{i}^{2}]$
$300 \times 2 = V_{f}^{2} - (5)^{2}$
$V_{f}^{2} = 600 + 25, V_{f} = 25$

Q. The acceleration vs distance graph for a particle moving with initial velocity 5m/s is shown in the figure. The velocity of the particle at x=35m will be