A bus was moving with a speed of 54km/h. On applying brakes, it stoppe

1.	A bus was moving with a speed of 54km/h. On applying brakes, it stopped in 8 seconds. Calculate the acceleration and distance travelled before stopping.Please solve it .
Answer» GIVEN that, the bus is moving with initial velocity (u) = 54 km/hr = 15 m/s On applying BRAKES, it stopped in 8 seconds. Final velocity (V) = 0 km/hr and Time = 8 sec We have to find the acceleration and DISTANCE travelled before stopping. i) For acceleration: According to Newton's First Law of Motion v = u + at → 0 = 15 + a(8) → -15 = 8a → -15/8 = a → a = -1.875 m/s² (Negative signs shows deacceleration) ii) For distance According to Newton's Second Law of Motion s = ut + 1/2 at² → s = 15(8) + 1/2 × (-1.875)(8)² → s = 120 + (-120/2) → s = 120 - 60 → s = 60 m

A bus was moving with a speed of 54km/h. On applying brakes, it stopped in 8 seconds. Calculate the acceleration and distance travelled before stopping.Please solve it .

Answer»

GIVEN that, the bus is moving with initial velocity (u) = 54 km/hr = 15 m/s

On applying BRAKES, it stopped in 8 seconds.

Final velocity (V) = 0 km/hr and Time = 8 sec

We have to find the acceleration and DISTANCE travelled before stopping.

i) For acceleration:

According to Newton's First Law of Motion

v = u + at

→ 0 = 15 + a(8)

→ -15 = 8a

→ -15/8 = a

→ a = -1.875 m/s²

(Negative signs shows deacceleration)

ii) For distance

According to Newton's Second Law of Motion

s = ut + 1/2 at²

→ s = 15(8) + 1/2 × (-1.875)(8)²

→ s = 120 + (-120/2)

→ s = 120 - 60

→ s = 60 m

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