Question

Seeing a boy waving hands, a taxi driver decreases its speed from 81 km/h to 1 km/h in 5 s.
Find the acceleration of the taxi.
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Answer 1

$\large\bf \underline\red{Question:-}$

Seeing a boy waving hands, a taxi driver decreases its speed from 81 km/h to 1 km/h in 5 s.

Find the acceleration of the taxi.

$\large\bf \underline\red{Answer:-}$

Initial speed of bus = u = 80km/ hour

=80$\times 1000$ ÷ 3600 s

800 / 36 m^-1

Final speed of bus =v= 60km / hour

=60×1000m ÷ 3600 s

= 600÷ 36 ms^-1= 600/36 ms^ -1

Therefore,

Acceleration,

a= v-u/t

600/36-800/36 ÷ 5

200 / 36×5 ms^-2

-10/9 ms^-2

-1.11 ms^-2

So the bus decelerating at a rate of -11.1 m/s^2

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Answer 2

Answer:

acceleration = -16Km/sec

Explanation:

• Acceleration is rate of change in velocity                                                                                          

• Here change in velocity is 1 - 81 = - 80 (final - initial)                                                              

• To get the rate we should divide the #change in velocity with time                                                                  
• Here time is 5 sec                                                                ∴  acceleration(Rate of change of Velocity) =  change in  velocity ÷ time                                                    
• = - 80 ÷ 5 = - 16 ( negative acceleration is called deceleration)
• So we tell the car is decelerating.