Question

with a speed of 54km/hr .on applying brakes it stopped in a bus was moving 8 seconds. calculate the acceleration and the distance travelled before stopping
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Answer 1

Answer:

initial velocity==54km/hr=(54×5/18)m/s=15m/s

final velocity =v=0 m/s

time=t=8sec

acceleration =a=(v-u)/t

a=(0-15)/8

a=(-15/8)= -1.875m/s²

Using eqn of motion

S=ut+1/2at²

S= 15×8+1/2×-1.875×8×8

S=120-60

S=60m

Therefore distance traveled before coming to rest is 60m.

Answer 2

$Question$

With a speed of 54km/hr .on applying brakes it stopped in a bus was moving 8 seconds.Calculate the acceleration and the distance travelled before stopping?

$Answer$

According to the question,

Initial Velocity of bus = 54km/hr = 54 x 5/18 = 15m/s

Final Velocity of bus = 0 (As break are applied the bus will come to rest )

Time taken to reach the Final Velocity = 8 sec

$Acceleration = \frac{v - u}{t}$

So, a = 0 - 15/8

Retardation = -1.875 m/s²

For calculating distance,

S = u x t + 1/2 at²

S = ( 15 x 8 ) + 1/2 x -1.875 + 8²

S = 120 + (-120/2)

S = 120 + ( -60)

S = 120 - 60

S = 60

Distance travelled by the bus before stopping = 60 m