Question

A car is moving with a speed of 50 metre per second brakes are applied and car comes to rest after travelling 3 metre find out
1.) time taken to coming rest.
2.) Distance travelled by car after applying when the speed is doubled. ​

Answer 1

Answer :

➡ Time taken by the car to stop after applying brakes = 3 minutes

➡ distance travelled when speed is doubled before applying brakes = 300 m

Step by Step explanations :

given that,

a car moving a speed of 50m/s

here,

initial velocity of the car = 50 m/s

also. given that,

it applies brakes and comes to rest in 3m,

here,

we have,

final velocity of the car = 0 m/s [car stopped]

time taken by the car to stop after applying brakes = 3 minutes

now we have,

initial velocity(u) = 50 m/s

final velocity(v) = 0 m/s

time taken(t) = 3 minutes

so,

by the equation of motion

v = u + at

where a is the acceleration of the car

putting the values,

0 = 50 + a(3)

3a = -50

a = -50/3 m/s

NOTE: negation of acceleration shows the retardation.

now,

given initial velocity(u) = 100 m/s

at the same retardation of the same car

here, we have,

initial velocity(u) = 100 m/s

acceleration(a) = -50/3 m/s²

final velocity(v) = 0 m/s [car will stop]

by the equation of motion

v² = u² + 2as

where,. S = distance travelled

putting the values,

(0)² = 100² + 2(-50/3)s

-100s/3 + 10000 = 0

-100s/3 = -10000 = 0

S = -10000 × 3/-100

S = 300 m

so,

distance travelled by the car after applying brakes when initial velocity of the car will 100 m/s

= 300 m

_______________

Answer :

➡ Time taken by the car to stop after applying brakes = 3 minutes

➡ distance travelled when speed is doubled before applying brakes = 300 m

Answer 2

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Answer :

• ➡ Time taken by the car to stop after applying brakes = 3 minutes

• ➡ distance travelled when speed is doubled before applying brakes = 300 m

Step by Step explanations :

given that,

a car moving a speed of 50m/s

here,

initial velocity of the car = 50 m/s

also. given that,

it applies brakes and comes to rest in 3m,

here,

we have,

final velocity of the car = 0 m/s [car stopped]

time taken by the car to stop after applying brakes = 3 minutes

now we have,

initial velocity(u) = 50 m/s

final velocity(v) = 0 m/s

time taken(t) = 3 minutes

so,

by the equation of motion

v = u + at

where a is the acceleration of the car

putting the values,

0 = 50 + a(3)

3a = -50

a = -50/3 m/s

NOTE: negation of acceleration shows the retardation.

now,

given initial velocity(u) = 100 m/s

at the same retardation of the same car

here, we have,

initial velocity(u) = 100 m/s

acceleration(a) = -50/3 m/s²

final velocity(v) = 0 m/s [car will stop]

by the equation of motion

v² = u² + 2as

where,. S = distance travelled

putting the values,

(0)² = 100² + 2(-50/3)s

-100s/3 + 10000 = 0

-100s/3 = -10000 = 0

S = -10000 × 3/-100

S = 300 m

so,

distance travelled by the car after applying brakes when initial velocity of the car will 100 m/s

= 300 m

_______________

Answer :

• ➡ Time taken by the car to stop after applying brakes = 3 minutes

• ➡ distance travelled when speed is doubled before applying brakes = 300 m.

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