Question

A particle has initial velocity of 17 ms-1

towards east and constant acceleration of 2ms-2

due west. The

distance covered by it in 9 th second of motion is :​

Answer 1

Answer:

1/2 m

0 m

Explanation:

A particle has initial velocity of 17 ms-1  towards east and constant acceleration of 2ms-2 due west. The distance covered by it in 9 th second of motion is :​

Initial Velocity = 17 m/s

Acceleration = 2 m/s² West = -2 m/s² East

Velocity after 8 sec = 17 -2*8 = 1 m/s  Towards East

Velocity after 9 Sec = 17 - 2*9 = -1 m/s => 1 m/s Towards West

So Particle direction changes between 8 to 9 sec

Velocity become = 0  at 17/2 = 8.5 Sec

Distance between 8 to 8.5 sec

0² - 1² = 2*(-2)S

=> S = 1/4 m

Distance between 8.5 to 9 sec

= (-1)² - 0² = 2(-2)S

=> S = -1/4 m

Distance Covered = 1/4 + 1/4 = 1/2 m

Displacement = 1/4 - 1/4 = 0 m