Question

a car travels with a uniform velocity of 55ms-1 for 10s. the breaks are then applied and the car is uniformly retarded and comes to rest in further 20s find the distance which the car travels before the brakes are applied and the retardation




Pls pls answer fast and I’ll mark the answer as the brainliest

Answer 1

Given :-

• A car travels with a uniform velocity of 55ms-1 for 10s. the breaks are then applied and the car is uniformly retarded and comes to rest in further 20s

To find :-

• The distance which the car travels before the brakes are applied and the retardation

Solution :-

• Speed = 55m/s

• Time = 10s

As we know that

→ Distance = speed × time

→ Distance = 55 × 10

→ Distance = 550m

Hence, distance which the car travels before the brakes are applied is 550m

• Initial velocity (u) = 55m/s

• Final velocity (v) = 0 (brakes applied)

• Time (t) = 20s

According to the first equation of motion

→ v = u + at

Put the value of v, u and t

→ 0 = 55 + a × 20

→ 20a = - 55

→ a = -55/20

→ a = - 2.75m/s²

• Minus shows retardation of car

Therefore,

• Retardation of car is 2.75m/s²

More to know :-

• v² - u² = 2as (Third equation of motion)

• s = ut + ½ at² (Second equation of motion)

Answer 2

Given :-

• Speed = 55m/s.

• Time with uniform velocity = 10 seconds.

• Time when car is retarded and has come to rest = 20 seconds.

• Initial Velocity = 55m/s.

• Final Velocity = 0m/s.

To Find :-

• Distance Traveled and Retardation.

Solution :-

We know that,

Distance = Speed × Time.

Put the values.

⇢ Distance = 55 × 10.

⇢ Distance = 550m.

Now, Calculating Acceleration.

v = u + at.

Put the values.

⇢ 0 = 55 + 20a.

⇢ 20a = -55.

⇢ a = -55/20.

⇢ a = -2.75 m/s²

Hence, These are your required answers.