Question

a car moving with a speed of 54 kilometre per hour is brought to rest within 10 seconds by applying brakes find redtatation its of the car​

Answer 1

Answer:

1.5 m/s² (retardation)

Explanation:

Given:

• Initial velocity of the car = u = 54 km/h

54 km/h = $54 \times \frac{5}{18}$= 15 m/s

• Final velocity of the car = v = 0 m/s (As it is brought to rest finally)
• Time taken = 10 seconds

To find:

• Acceleration of the car

Acceleration = $\dfrac{Final \ velocity - Initial \ velocity}{Time}$

Acceleration = $\frac{0-15}{10}$

Acceleration = $\frac{-15}{10}$

Acceleration = -1.5 m/s²

The retardation of the car is equal to 1.5 m/s²

Note : Here acceleration will be -1.5 m/s² but retardation will be 1.5 m/s²

Answer 2

✰ Given:-

▪Initial velocity (u) =54×5/18=15m/s

▪ Final velocity (v) =0m/s

▪Time taken (t) =10s

✰To Find:-

▪ Retardation of car.

✰Solution :-

By using first equation of motion

➠ v=u+at

➠ 0=15+a×10

➠ a =-15/10

➠ a=-3/2

➠ a= -1.5ms^2.

∴The retardation of car is 1.5m/s^2.

✰Additional information ✰

➧Negative acceleration are called retardation.

➧ SI unit is m/s^2

➧ It is vector quantity.