Question

A bus decreases its speed from 80 km per hour from 60 km per hour in 5 seconds calculate the acceleration and distance

Answer 1

Answer:

a = -1.112 m/s²

DISTANCE =97.2m

Explanation:

80 km/hr = 200/9 m/s = u

60 km/hr = 100/6 m/s = v

a = (v - u)/t

a = (100/6 - 200/9)/5

a =(16.66 - 22.22)/5

a =  -5.56/5 m/s² = -1.112m/s²

S = ut + 1/2 at²

S = 22.22 x 5 +  1/2 (-1.112) x 5²

S= 111.1 - 13.9

S= 97.2 m

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Answer 2

Question

A bus decreases its speed from 80 km per hour to 60 km per hour in 5 seconds. Calculate the acceleration and distance covered by the bus.

Solution-

A bus decreases its speed from 80 km per hour to 60 km per hour in 5 seconds.

Here, the initial speed of the bus is 80 km/hr and final speed of the bus is 60 km/hr and time taken by the bus is 5 sec.

We have to find the acceleration and distance covered by the bus.

Firstly convert km/hr into m/s. To do this, multiply with 5/18.

Initial speed/velocity (u) = 80 × 5/18 = 22.22 m/s

Final speed/velocity (v) = 60 × 5/18 = 16.67 m/s

Using the First Equation Of Motion,

v = u + at

Substitute the known values in the above formula,

16.67 = 22.22 + a(5)

16.67 - 22.22 = 5a

-5.55 = 5a

-1.11 = a

(Negative sign shows retardation)

Therefore, the acceleration of the bus is 1.11 m/s².

Now, using the Second Equation Of Motion,

s = ut + 1/2 at²

Substitute the known values,

s = 22.22(5) + 1/2 × (-1.11)(5)²

s = 111.1 - 1.11/2 × 25

s = 111.1 - 0.555 × 25

s = 111.1 - 13.875

s = 97.225

Therefore, the distance covered by the bus is 97.23 m.