Question

A bus moving at a speed of 72 km/hr . on applying brakes it come to a rest in 5 sec . find the acceleration and Distance covered by the bus.​

Answer 1

Given:-

initial velocity = 72km/hr = 72 × 5/18 m/s = 20 m/s

final velocity = 0 m/s

time = 5 sec

To Find:-

Acceleration = ?

Distance = ?

Solution:-

By using 1st kinematical equation:-

$\\\bullet\quad\underline{\boxed{\sf v = u+at }}\bigstar$

Substitute the values,

⇒ v = u + at

⇒ 0 = 20 + a(5)

⇒ 0 = 20 + 5a

⇒ a = - 20/5

⇒ a = - 4 m/s²

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By using 3rd kinematical equation:-

$\\\bullet\quad\underline{\boxed{\sf v^{2}-u^{2} = 2as }}\bigstar$

Substitute the values,

⇒ (0)² - (20)² = 2(-4)s

⇒ 0 - 400 = -8s

⇒ -8s = -400

⇒ s = 400/8

⇒ s = 50m

∴ The required answers are -4m/s² and 50m

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