Question

A bus moves with initial velocity of 36 km/h and acceleration 60 cm/s 2 over a distance of 500 m. The final velocity of bus is?​

Answer 1

Given ::

Initial velocity = 36km/h ( 10m/s )

Acceleration = 60 cm s^-2 ( 0.6 m s^-2 )

Distance covered with acceleration = 500 m

( v - final speed , u - initial speed, a - acceleration , s = distance covered )

Calculations ::

Using v^2 - u^2 = 2as

v^2 = 2as + u^2

= 600 + 100

= 700 m/s

On square rooting both sides

v = √700

v = 26.457 m/s

On converting it into km/h

= 96.257 km/hr

Final velocity = 96.257 km/hr

Extra information ::

• To convert m/sec into km/hr multiply by 18 and divide by 5

• To convert km/hr into m/sec multiply by 5 and divide by 18

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Answer 2

The Final Velocity Of The Bus Is 26.4 m/s

GIVEN

The initial velocity of the bus = 36 km/hr

Acceleration of the bus = 60 cm/s^2

Distance covered = 500 meters.

TO FIND

The final velocity of the bus.

SOLUTION

We can simply solve the above problem as follows-

We are given,

The initial velocity of the bus = 36 km/hr

Acceleration of the bus = 60 cm/s^2

Distance covered = 500 meters

To calculate the final velocity of the bus, we will first convert the given values into the same units.

Initial velocity =

$36 \times \frac{5}{18} = 10m{s}^{ - 1}$

Acceleration of the bus =

$60 \times 0.01 = 0.6m {s}^{ - 2}$

Distance = 500 meters.

Now, we will apply the second equation of motion, which is as follows-

${v}^{2} - {u}^{2} = 2as$

Where,

v = final velocity

u = initial velocity

a= acceleration

s= distance

Putting the values in the above equation we get

${v}^{2} - {10}^{2} = 2 \times 0.6 \times 500$

${v}^{2} = 600 + 100$

${v}^{2} = 700$

v = √700

v = 26.4

Hence, the final velocity of the bus is 26.4 m/s

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