Question

A car moving with a velocity of 40 ms' is stopped by the applcation of brakes which impact a retardation of 5 ms 2 to car : (a) How long does it take for come to a stop ? (b) How far does the car travel during the brakes are applied?​

Answer 1

$\Large {\underline { \bf {Solution :}}}$

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

• Initial velocity (u) = 40 m/s
• Final velocity (v) = 0 m/s [It stops]
• Acceleration (a) = - 5 m/s²

We have to calculate time taken by it for come to a stop. And, distance covered by the car.

By using the first equation of motion,

❍ ｖ ＝ ｕ ＋ ａｔ

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

→ 0 = 40 + (-5)t

→ 0 = 40 - 5t

→ - 40 = - 5t

→ (-40) ÷ (-5) = t

→ 8 s = t

Therefore, it travels for 8 seconds for come to a stop.

━━━━━━━━━━━━━━━━━━━━━━━━━━━

Now, by using the third equation of motion :

❍ ｖ² － ｕ² ＝ ２ａｓ

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

→ (0)² － (40)² ＝ 2 × (-5) × s

→ 0 － 1600 ＝－10 s

→ (－1600 ) ÷ (－10 ) = s

→ 160 m = s

Therefore, it travels 160 m during the brakes are applied.