Question

3. If a car moving a speed of 50m/s applies brakes and comes to rest in 3m, how long will the same car travel before coming to rest if its initial speed is 100m/s. ​

Answer 1

Answer:

18 km

Explanation:

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,

= 3 × 60 s

= 180 s

here,

we have,

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

time taken by the car to stop after applying brakes = 180 seconds

now we have,

initial velocity(u) = 50 m/s

final velocity(v) = 0 m/s

time taken(t) = 60 seconds

so,

by the equation of motion

v = u + at

where a is the acceleration of the car

putting the values,

0 = 50 + a(180)

180a = -50

a = -50/180 m/s

a = -5/18 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) = -5/18 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(-5/18)s

-10s/18 + 10000 = 0

-10s/18 = -10000

S = -10000 × 18/-10

S = 18000 m

so,

S = 18 km

so,

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

= 18 km