Question

A particle is moving with constant acceleration 3m/s due west has an initial velocity of 2m/s due east find distance covered in fith second of its motion ? answer with problem

Answer 1

dcgx guhv bhihcfbj jygoo boink v kg c. lgv

Answer 2

QUESTION - A particle is moving with constant acceleration 3m/s due west has an initial velocity of 2m/s due east find distance covered in fith second of its motion ?

ANSWER ------

Initial velocity (u) = 2 m/sec towards east
time (t) = 5 sec
Acceleration (a) = (-3) m/sec towards east = 3 m/sec towards west


USING SECOND LAW OF MOTION

distance covered (s) = ut + 1/2 a (t square)

s = 2 × 5 + 1/2 × (-3) × 5 × 5
s = 10 + -75/2
s = (20 - 75) / 2
s = -55 / 2

s = -27.5 m

Since distance covered can never be negative therefore distance covered = + 27.5 m

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
ANSWER = 27.5 m
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>