Question

A bus is moving with a speed 72 km/h can be stopped by brakes
after atleast 10 m. What will be the minimum stopping distance, if the
same bus is moving at a speed of 144 km/h?
​

Answer 1

Answer:

20 metres

Explanation:

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Answer 2

Given -

• Initial velocity = 72 km/hr

• Distance = 10m

• Final Velocity = 0 m/s

To find -

• Minimum stopping distance, when speed is 144 km/hr

Formula used -

• 3rd equation of motion.

Solution -

In the question, we are provided with the distance covered, initial and final Velocity, and we need to find the minimum stopping distance, for that first we are going to find the acceleration, by using the 3rd law of motion, and then we will find the minimum stopping distance, by again using, 3rd law of motion. Let's do it!

So -

$\sf \underline{3rd \: law \: of \: motion} \: = {v}^{2} \: - {u}^{2} \: = 2as$

Where -

v = Final velocity = 0 m/s

u = Initial velocity = 72 km/hr = 20 m/s

a = Acceleration

s = Distance

On substituting the values-

$\sf \longrightarrow \: {0}^{2} - {20}^{2} = 2 \times a \times 10 \\ \\ \sf \longrightarrow \: - 400 = 20a \\ \\ \sf \longrightarrow \: a = \cancel \dfrac{ - 400}{20} \\ \\ \sf \implies \: a = - 20m/s^{2}$

Now -

We will find the minimum stopping distance of bus, by again applying, 3rd law of motion, this time, we have, acceleration, so, we will apply it, and will find the distance.

$\sf \underline{3rd \: law \: of \: motion} \: = {v}^{2} \: - {u}^{2} = 2as$

Given that -

Acceleration = -20 m/s²

Final Velocity = 0 m/s

Initial velocity = 144 km/hr = 40 m/s

On substituting the values -

$\sf \longrightarrow{(0)}^{2} - {(40)}^{2} = 2 \times ( - 20) \times s \\ \\ \sf \longrightarrow \: 0 - 1600 = - 40 \times s \\ \\ \sf \longrightarrow \: - 1600 = - 40 \times s \\ \\ \sf \longrightarrow \: s = \dfrac{ - 1600}{ - 40} \\ \\ \sf \implies \: s = 40m$

$\therefore$ The minimum distance, the bus can be stopped is 40m

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