Question

14. A particle is moving on a straight line path with initial velocity 45 m/s and along positive X-axis acceleration of 10 m/s along negative x-axis. The distance travelled by the particle in 5th second
motion is
(1) 2.5 m
(2) 30 m
(3) 180 m
(4) 0​

Answer 1

answer : option (1) 2.5 m

explanation : given,

Initial velocity of particle, u = 45m/s along positive direction of x-axis

acceleration of particle , a = 10 m/s² along negative direction of x-axis.

here we see, acceleration is just opposite in direction of velocity.

at a specific time , velocity of particle will be zero after then, motion of particle will be along acceleration.

final velocity, v = 0

using formula, v = u + at

so, 0 = 45 - 10t

or, t = 4.5sec

hence, acceleration and velocity are in opposite direction upto 4.5sec after that acceleration and velocity are in same direction.

so, distance travelled by particle in 5th second = distance travelled by particle during (4 to 4.5sec) + distance travelled by particle ( 4.5 to 5sec)

so, let's find velocity at 4sec [ it will be initial velocity for 4 to 4.5 sec]

U = u + at

= 45 + (-10)(4)

= 5 m/s

now, distance travelled by particle in 5th second = |Ut + 1/2at² | + | vt + 1/2 at² |

= | 5 × 0.5 + 1/2(-10)(0.5)² | + | 0 × 0.5 + 1/2 (-10)(0.5)² |

= | 2.5 - 1.25 | + |-1.25|

= 1.25 + 1.25

= 2.5m