Question

A particle starting with certain initial velocity and uniform acceleration covers a distance of 12 metre in first 3 seconds and a distance of 30 metre in next 3 seconds. The initial velocity of the particle is ?

Motion in a straight line. ​

Answer 1

Answer:

1 m/sec

Formula Used:

S = ut + 1/2 at²

Explanation:

Let u be the initial velocity that have to find and a be the uniform acceleration of the particle.

For t = 3s, distance travelled S = 12 m and for t = 3 + 3 = 6 s distance travelled Sc = 12 + 30 = 42m

From S = ut +1/2 at²

12 = u x 3 + 1/2 x a x 32

or 24 = 6u + 9a .....(i)

Similarly,  42 = u x 6 + 1/2 x a x 62

or 42 = 6u + 18a .....(ii)

On solving equation (i) & (ii), we get u = 1 m/sec

Conclusion:

The initial velocity of the particle is 1 m/sec

Answer 2

ok....

trying to solve.

let the initial velocity of the particle be u m/s.

the acceleration be a m/s²

we know that...s=u +½at²

then for the first case,

12= u+½a(3)²

or, a=(24-2u)÷9

for the second case ,

30= v1 + ½a(3)² (here v1 is the terminal velocity after 12 m)

or, v1=30-9a/2

or, v1=30-9(24-2u)/9×2

or, v1=30-12+u

or, v1=18+u

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.

. again we know V = u+ at

then, v1 = u+a(3)

. . . . .or, v1 = u+ 3a

now putting the value of v1 we get ,

18+u=u+3a

or, a= 6 m/s²

hence a =(24-2u)/9

. . . . .or, u=(-15)

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.

.

I don't know why it is getting negative....sorry.

please check it and make it right.