Physics, asked by satyarth09, 8 months ago

A particle has an initial velocity of (3ỉ+4) m/s and a constant acceleration of (4ỉ - 3)) m/s². Its speed after
one second will be equal to:​

Answers

Answered by akibkhansaif
0

Answer:

Initial velocity of particle

u

= 3

i

^

+4

j

^

Acceleration

a

= 4

i

^

−3

j

^

Final velocity after 1 second

v

=

u

+

a

t

where t=1 s

v

= (3

i

^

+4

j

^

)+ (4

i

^

−3

j

^

)×1= 7

i

^

+

j

^

Speed after 1 second ∣

v

∣=

7

2

+1

=

50

m I hope you this answer help to you and all of you

Answered by Ironman571y
0

Explanation:

using 1st equation of motion .

v=u+at,

placing the values:-

v=(3i+4)+(4i-3)×1

v=3i+4+4i-3

v=7i+1

so the final velocity = 7i+1

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