Question

find declaration of the car wich changes its velocity 33m/s to11m/ s in 10 seconds ​

Answer 1

Answer:

-2.2 m/s²

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 33 m/s
• Final velocity (v) = 11 m/s
• Time taken (t) = 10 seconds

We are asked to calculate deceleration of the car.

In order to calculate deceleration of the car, firstly we need to find the acceleration of the car.

As we know that,

$\implies \sf {a = \dfrac{v - u}{t} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\implies \sf {a = \Bigg ( \dfrac{33-11}{10} \Bigg ) \; ms^{-2}}$

$\implies \sf {a = \Bigg ( \dfrac{22}{10} \Bigg ) \; ms^{-2}}$

$\implies \sf {a = 2.2 \; ms^{-2}}$

$\implies \sf {Deceleration = -2.2 \; ms^{-2}}$

∴ Deceleration of the car is -2.2 m/s².

$\rule{200}2$

More Information :

• Acceleration is the rate of change in velocity.
• SI unit of acceleration is m/s².
• It is a vector quantity.
• Negative acceleration is called deceleration.
• Deceleration is also known as retardation.