Question

a car moving with a speed of 54km/h is brought to rest within 10 s by applying brakes . find the retardation due to brakes

Answer 1

Given :

▪ Initial velocity = 54kmph

▪ Final velocity = zero (i.e., rest)

▪ Time interval = 10s

To Find :

▪ Retardation due to brakes.

Concept :

↗ Acceleration is defined as the rate of change in velocity.

↗ It is a vector quantity.

↗ It has both magnitude as well as direction.

↗ It can be positive, negative or zero.

↗ SI unit : m/s²

Conversion :

➡ 1kmph = 5/18mps

➡ 54kmph = 54×5/18 = 15mps

CalculaTion :

✴ First equation of kinematics :

☞ v = u + at

☞ 0 = 15 + a(10)

☞ 10a = -15

☞ a = -1.5 m/s²

[Note : -ve sign shows retardation.]

Answer 2

Answer :

✴ Initial velocity = 54kmph = 15mps

✴ Final velocity = zero

✴ Time = 10s

____________________________________

→ Acceleration is defined as the ratio of change in velocity to the time interval.

• Vector Quantity
• SI unit : m/s²

Mathematically,

$\bigstar\:\underline{\boxed{\bf{\blue{a=\dfrac{v-u}{t}}}}}$

⭐ a denotes acceleration

⭐ v denotes final velocity

⭐ u denotes initial velocity

⭐ t denotes time

⏏ a = (v - u)/Δt

⏏ a = (0 - 15)/10

⏏ a = -15/10

⏏ a = -1.5 m/s² ✔

⚠ Negative sign of a indicates deceleration.