Question

An automobile of mass M accelerates from rest with a constant power P generated by the engine. Assume that 80 percentage of power generated by the engine accelerates the automobile.what is the distance covered by automobile?

Answer 1

Answer:

Correct option is

C

45

32

M

Pt

3

Let the mass of the body be M and P be the constant power supplied

Now , P=F.v , where F is the force applied by the source

Hence ,

P=M×a×v , where a is the acceleration of the body .

But a=

dt

dv

Hence ,

P=M×

dt

dv

×v

i.e,

10

8P

=F.v=M

dt

dv

v

Or

P.dt=Mv.dv

i.e,

10

8P

dt=Mv.dv

Integrating both sides from 0 to t and velocity from 0 to v , we get

Pt=

2

Mv

2

i.e,

5

4P

t=

2

Mv

2

v=

5M

8Pt

dt

dx

=

5M

8Pt

x=

5M

8Pt

dt

x=

5M

8P

(

3

2t

3/2

)

x=

3

2

5M

8P

×t

3/2

Hence , x=

45M

32Pt

3

is the distance covered by automobile in time t

Explanation:

HOPE IT HELPS FROM MY SIDE!

Answer 2

Answer:

Let the mass of the body be M and P be the constant power supplied

Now , P=F.v , where F is the force applied by the source

Hence ,

P=M×a×v , where a is the acceleration of the body .

But a=

dt

dv

Hence ,

P=M×

dt

dv

×v

i.e,

10

8P

=F.v=M

dt

dv

v

Or

P.dt=Mv.dv

i.e,

10

8P

dt=Mv.dv

Integrating both sides from 0 to t and velocity from 0 to v , we get

Pt=

2

Mv

2

i.e,

5

4P

t=

2

Mv

2

v=

5M

8Pt

dt

dx

=

5M

8Pt

x=

5M

8Pt

dt

x=

5M

8P

(

3

2t

3/2

)

x=

3

2

5M

8P

×t

3/2

Hence , x=

45M

32Pt

3