Physics, asked by raziipadmapati, 10 days ago

a body initially at rest starts moving with a constant acceleration 2m/s
Calculate- velocity aquired
distance covered in 5secs

Answers

Answered by raj3294
2

Answer:

FINAL VELOCITY ,V = 10 M/S

DISTANCE TRAVELLED , S= 25 M

Explanation:

STARTS FROM REST INDICATES,

INITIAL VELOCITY ,U = 0 M/S

ACCELERATION,A = 2 M/S²

TIME TAKEN , T = 5 SEC

DISTANCE TRAVELLED ,S=?

FINAL VELOCITY,V =?

ACCORDING TO NEWTON'S SECOND LAWS OF MOTION

S = U T +1/2 A T²

S = 0*5 + 1/2*2*5*5

S = 0+25

S = 25 M

V = U+A T

V = 0+2*5

V = 10 M/S

THEREFORE FINAL VELOCITY IS 10 M/S.

THEREFORE THE DISTANCE TRAVELLED IS 25 M.

HOPE THIS HELPS.

Answered by shubhamchaudhary8305
2

Answer:

Final velocity=10 m/sec

Distance covered= 25 m

Explanation:

According to the first equation of motion

v=u+at

Here, u= 0 m/sec

         a= 2 m/sec²

         t = 5 seconds

v= 0+2×5= 0+10= 10 m/sec

According to the third equation of motion

v²-u²=2 as

(10)²-(0)²=2×2×s

100-0=4s

100/4=s

s=25m

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