A particle covers 10m distance in 2sec. if final velocity is 8m/s, find initial velocity​

Explanation:

It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion

Case I

s=ut+

2

1

at

2

10=5u+

2

1

a(5)

2

20=10u+25a

4=2u+5a(1)

Case 2

In next 3s the particle covers more 10m distance. So

20=8u+

2

1

a(8)

2

5=2u+8a(2)

On solving equation (1) and (2)

4=2u+5a

5=2u+8a

a=

3

1

m/s

2

Put the value of a in equation (1)

u=

6

7

m/s

Now to find distance in next 10 s. total time will be 10s

s=

6

7

×10+

2

1

×

3

1

×(10)

2

s=28.33m

Distance travelled in next 2 sec

s=28.33−20=8.33m

vi =2

Explanation:

Δx = 1/2 (vi + vf) t

10 = 1/2 ( vi + 8 ) 2

10 = vi + 8

10-8 = vi

vi = 2 m/s

vi= initial velocity

vf= final velocity

Δx= distance

t= time

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