Physics, asked by sanjusrinu007, 11 months ago

Engine of a car supplies constant power to the car which accelerates from rest. Car moves along a straight road..

Attachments:

Answers

Answered by samyuktha0025
24

Answer: 3rd option.

Explanation :

I think it is the correct way. If you get a simpler way to do this please do inform me.

Attachments:
Answered by tiwariakdi
0

Answer:

The correct option is C velocity of the car is proportional to

t

1

/

2

.

P

=

F

×

v

=

m

a

×

v

=

m

v

t

×

v

P

t

m

=

v

2

v

t

In rear-wheel drive cars, the engine rotates the rear wheels and the front wheels rotate only because the car moves. If such a car accelerates on a horizontal road, the friction

Explanation:

In rear-wheel drive cars, the engine rotates the rear wheels and the front wheels rotate only because the car moves. If such a car accelerates on a horizontal road, the friction

If t

1

 be the time for which the car accelerates at the rate f from rest, the distance traveled in time t

1

 is

s=s

1

=0(t

1

)+

2

1

ft

1

2

=

2

1

ft

1

2

 ....(1) (using formula

s=ut+

2

1

at

2

)

and the velocity at time t

1

 is v

1

=ft

1

So, the distance traveled in time t will be s

2

=v

1

t=ft

1

t ...(2)

If t

2

 be the time for which the car decelerates at the rate f/2 to come rest, the distance traveled in time t

2

 is given by,

0

2

−v

1

2

=2(−f/2)s

3

   (using formula v

2

−u

2

=2as)

or s

3

=(ft

1

)

2

/f=ft

1

2

 ...(3)

using (1),  s

3

=2s

1

=2s

Given, s

1

+s

2

+s

3

=15s

or s+ft

1

t+2s=15s

or ftt

1

=12s

or 12s=ftt

1

 ..(4)

(4)/(1)⇒

s

12s

=

(1/2)ft

1

2

ftt

1

 or t

1

=t/6

From (1), s=(1/2)f(t/6)

2

=ft

2

/72

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