Question

A 1300 kg car was traveling 45 m/s towards a traffic signal when the light suddenly turned red. The driver slammed on her brakes and stopped the car in 3.3 seconds. What was the deceleration of the car?

Answer 1

★ Acceleration (a) : It is defined as rate of change in velocity

➳ a = ( v - u ) / t $\;$

➳ SI units : m/s²

★ Deceleration : Acceleration and deceleration/retardation are both same but opposite in direction .

➳ Deceleration = - Acceleration

Mass of the car (m) = 1300 kg

Initial velocity of the car (u) = 45 m/s

While travelling towards traffic signal when the light suddenly turned red , the driver slammed on her brakes . So ,

Final velocity of the car (v) = 0 m/s

Time taken to stop (t) = 3.3 s

➠ a = ( v - u ) / t

➠ a = ( 0 - 45 ) / 3.3

➠ a = - 13.63 m/s²

Note : - ve sign of acceleration denotes retardation / deceleration

★ Deceleration = 13.63 m/s²

Answer 2

Answer:

• The deceleration of the car is 13.63 m/s².

Explanation:

Given that,

• Mass (m) = 1300 kg
• Initial velocity (u) = 45 m/s
• Final velocity (v) = 0 m/s
• Time (t) = 3.3 sec

As we know that,

↪ Deceleration = - Acceleration [ °.° Acceleration and deceleration are both same but opposite site in direction.]

❤ Applying 1st eqn. of motion, ❤

↪ v = u + at

[ Putting values ]

➡ 0 = 45 + a × 3.3

➡ -45 = 3.3a

➡ a = -45/3.3

➡ a = -13.63 m/s²

[ Negative sign of acceleration denote's deceleration ]

$\red\star$ Deceleration = 13.63 m/s² $\green\star$