Question

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 .

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Answer 1

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

Answer 2

Given:-

The bus had an initial velocity of 54 km/h.

On applying brakes, the bus took a time of 8 secs to stop.

To calculate:-

The acceleration of the bus before stopping.

Solution:-

Initial velocity of the bus (u) = 54 km/h = 54/3.6 m/s = 15 m/s

Final velocity of the bus (v) = 0 km/h = 0 m/s   (∵As the bus finally stops)

Using Newton's first law of motion,

v = u + at

⇒0 = 15 + 8a

⇒ -15 = 8a

⇒a = -15/8

⇒$\boxed{a=-1.875\ m/s^2}$

As the sign is negative, so a deceleration has occurred.

Using Newton's second law of motion,

s = ut + 1/2 at²

⇒s = 15*8 + 1/2*-1.875*(8)²

⇒s = 120 + 1/2*-120

⇒s = 120 + (-60)

⇒s = 120 - 60

⇒$\boxed{s=60\ m}$

∴So, the acceleration and the distance of the bus before stopping are -1.875 m/s² and 60 m respectively.