Velocity of a particle at time t =0 is 2m/s . A constant acceleration

1.	Velocity of a particle at time t =0 is 2m/s . A constant acceleration of 2 m/s2 acts on the particle for 2 sec at an angle of 60 with its initial velocity . find it magnitude of velocity and displacement of the particle at the end of t = 2s ?
Answer» Answer: final velocity = 4m/s displacement = 12m Explanation: as given initial velocity of particle v=2m/s ACCELERATION of particle in direction of initial velocity a=(2×cos60)m/s^2 a=1m/sec^2 time given, t=2sec to find final velocity we use laws of motion v'=v+at v'=2+(1×2) v'=4m/s to find displacement we use s=vt+1/2a(t^2) s=(2×2)+(2×1×4) s=12m perpendicular component of acceleration balances the weight of the body so we have a VERTICAL equilibrium and it doesn't affect the HORIZONTAL motion of the body so we can IGNORE it

Velocity of a particle at time t =0 is 2m/s . A constant acceleration of 2 m/s2 acts on the particle for 2 sec at an angle of 60 with its initial velocity . find it magnitude of velocity and displacement of the particle at the end of t = 2s ?

Answer»

Answer:

final velocity = 4m/s

displacement = 12m

Explanation:

as given initial velocity of particle

v=2m/s

ACCELERATION of particle in direction of initial velocity

a=(2×cos60)m/s^2

a=1m/sec^2

time given, t=2sec

to find final velocity we use laws of motion

v'=v+at

v'=2+(1×2)

v'=4m/s

to find displacement we use

s=vt+1/2a(t^2)

s=(2×2)+(2×1×4)

s=12m

perpendicular component of acceleration balances the weight of the body so we have a VERTICAL equilibrium and it doesn't affect the HORIZONTAL motion of the body so we can IGNORE it

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