Question

the particle is moving along a straight line with initial velocity 5m/sec towards positive x-axis a constant acceleration of 2.5m/s² towards negative x axis starts acting on a particle at t=0.find A. time at which particle reverses direction.B.distance at 1sec,2sec,3sec,4sec,5sec.C. displacement at 1 sec,2sec,3sec,4sec,6sec.​

Answer 1

Answer:

Area under the a−t graph gives the change in the velocity because dv=adt

so the change in velocity i.e area is Δv=positive rectangle + positive triangle + negative big triangle

or Δv=5×1+

2

1

×(1.5−1)×5−

2

1

×(4−1.5)×10=−6.25m/s

so the new velocity will be v

i

+Δv=5−6.25=−1.25m/s

MARK AS BRANIEST