The acceleration of a particle traveling along a straight line is shown in the figure. The maximum speed of the particle is

Question

The acceleration of a particle traveling along a straight line is shown in the figure. The maximum speed of the particle is (A) 20 ms -1 (B) 30 ms -1 (C) 40 ms -1 (D) 60 ms -1

Tardigrade Admin · Accepted Answer

Maximum speed is at t=8 sec <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7ce695c54e950c96bb08469fb3990e06-.png" /> v at t=8 sec is given by are under curve O A B C. Max. speed =(1/2)(A B+O C) × B C =(1/2) ×(12) × 5=30 ms -1

Q. The acceleration of a particle traveling along a straight line is shown in the figure. The maximum speed of the particle is

$20 m s^{- 1}$

$30 m s^{- 1}$

$40 m s^{- 1}$

$60 m s^{- 1}$

Maximum speed is at $t = 8 sec$

$v$ at $t = 8 sec$ is given by are under curve $O A BC$ .
Max. speed $= \frac{1}{2} (A B + OC) \times BC$
$= \frac{1}{2} \times (12) \times 5 = 30 m s^{- 1}$

Q. The acceleration of a particle traveling along a straight line is shown in the figure. The maximum speed of the particle is

20ms−1

30ms−1

40ms−1

60ms−1

Maximum speed is at t=8sec v at t=8sec is given by are under curve OABC. Max. speed =21​(AB+OC)×BC =21​×(12)×5=30ms−1

$20 m s^{- 1}$

$30 m s^{- 1}$

$40 m s^{- 1}$

$60 m s^{- 1}$

Maximum speed is at $t = 8 sec$

$v$ at $t = 8 sec$ is given by are under curve $O A BC$ .
Max. speed $= \frac{1}{2} (A B + OC) \times BC$
$= \frac{1}{2} \times (12) \times 5 = 30 m s^{- 1}$