Question

A car of mass 200 kg is moving with a certain velocity. It is brought to rest by the application of brakes, within a distance of 20m when the average resistance being offered to it is 500N. What was the velocity of the car?​

Answer 1

◇QUESTION ◇:-

A car of mass 200 kg is moving with a certain velocity. It is brought to rest by the application of brakes, within a distance of 20m when the average resistance being offered to it is 500N. What was the velocity of the car?

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◇ANSWER ◇:- 10m/s

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◇SOLUTION ◇:-

GIVEN: - m = 200kg

s = 20m

F = - 500 N.

REQUIRED TO FIND: - Velocity of the car.

CONCEPT &FORMULA: -

》We know,

${\boxed{a = F/m}}$

where, a = acceleration; F = force ; m = mass

》And,

${\boxed{v {}^{2} - u {}^{2} = 2as }}$

where, v = final velocity & u = Initial velocity.

CALCULATION: -

》 a = F /m = -500/200

》a = -2.5 m/s^2.

Now,

$= > u {}^{2} = v {}^{2} - 2as$

=>= 0-2 (-2.5)× 20

=> = 100 m^2/s^2

=> u = 10 m/s

Hence, ${\boxed{velocity = 10m/s }}$

HOPE IT HELPS :)

Answer 2

Answer:

The energy present in electric current.

Explanation:

GIVEN: - m = 200kg

s = 20m

F = - 500 N.

REQUIRED TO FIND: - Velocity of the car.

CONCEPT &FORMULA: -

》We know,

{\boxed{a = F/m}}

a=F/m

where, a = acceleration; F = force ; m = mass

》And,

{\boxed{v {}^{2} - u {}^{2} = 2as }}

v

2

−u

2

=2as

where, v = final velocity & u = Initial velocity.

CALCULATION: -

》 a = F /m = -500/200

》a = -2.5 m/s^2.

Now,

= > u {}^{2} = v {}^{2} - 2as=>u

2

=v

2

−2as

=>= 0-2 (-2.5)× 20

=> = 100 m^2/s^2

=> u = 10 m/s

Hence, {\boxed{velocity = 10m/s }}

velocity=10m/s

HOPE IT HELPS :)