Physics, asked by Rashikasera, 1 year ago

Q18. A car moving with s speed of 40 m/s can be stopped by applying brakes in 2 m. If the same acceleration exists, after how much distance in which a car with velocity 80 m/s can be stopped?​

Answers

Answered by deepsen640
22

Answer:

when the velocity of car will 80 m/s

distance travelled = 8 m

Step by step explanations :

given that,

A car moving with s speed of 40 m/s

can be stopped by

applying brakes in 2 m.

here,

initial velocity of the car ⛟ = 40 m/s

final velocity of the car = 0 m/s

[it comes to rest]

distance travelled after applying brakes

= 2 m

let the retardation produced by the brakes be a

now we have,

initial velocity(u) = 40 m/s

final velocity(v) = 0 m/s

distance travelled(s) = 2 m

acceleration(a) = a

by the equation of motion,

v² = u² + 2as

putting the values,

(0)² = 40² + 2a(2)

4a + 1600 = 0

4a = -1600

a = -1600/4

a = -400

Negation of acceleration shows the retardation

now,

given initial velocity of the car = 80 m/acceleration is the same

= -400 m/s

final velocity(v) = 0 m/s

again

v² = u² + 2as

putting the values,

(0)² = 80² + 2(-400)s

-800s + 6400 = 0

-800s = -6400

S = -6400/-800

S = 8 m

so,

when the velocity of car will 80 m/s

distance travelled = 8 m

Answered by ILLIgalAttitude
24

Answer:

distance travelled = 8 m

Step by step explanations :

given,

for a car

initial velocity(u) = 40 m/s

final velocity(v) = 0 m/s

[brake applied to rest]

distance travelled(s) = 2 m

v² = u² + 2as

a = acceleration

0² = 40² + 2a(2)

4a = -1600

a = -1600/4

a = -400

initial velocity = 80 m/s

acceleration = -400 m/s

final velocity = 0 m/s

v² = u² + 2as

0² = 80² + 2(-400)s

-800s = -6400

S = 6400/800

S = 8 m

so,

distance travelled = 8 m

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