Question

A bus was moving with a speed of 54km/h. On applying brakes, it stopped in 5 seconds.

Calculate (a) the acceleration and (b) distance travelled before stopping.​

Answer 1

Answer:

Explanation:

Given,

Initial velocity, u = 54 km/hr = 54 × 5/18 = 15 m/s

Final velocity, v = 0 m/s (As breakes applied)

Time = 5 seconds

To Find,

a. Acceleration

b. Distance travelled

Formula to be used,

1st equation of motion, i.e, v = u + at

3rd equation of motion, i.e, v² - u² = 2as

Solution,

Putting all the values, we get

v = u + at

⇒ 0 = 15 + a × 5

⇒ - 15 = 5a

⇒ - 15/5 = a

⇒ a = - 3 m/s².

As the sign is negative, so deceleration occurred.

Now, the distance travelled,

v² - u² = 2as

⇒ (0)² - (15)² = 2 × (- 3) × s

⇒ 225 = - 6s

⇒ 225/6 = s

⇒ s = 37.5 m

Hence, the distance travelled by bus is 37.5 m.

Answer 2

Given:-

• A bus was moving with a speed of 54km/h. On applying brakes, it stopped in 5 seconds.

To Find:-

• The acceleration and distance travelled before stopping.

Equation Used:-

• ${\boxed{\bf{First\: Equation\: of\: Motion: v = u+at}}}$

• ${\boxed{\bf{Third\: Equation\: of\: Motion: v^2-u^2 = 2as}}}$

Solution:-

Using First Equation of Motion,

$\sf :\implies\:v = u +at$

Here,

• v = 0m/s

• u = 15m/s

• t = 5 sec

Putting values,

$\sf :\implies\:0 = 15+a\times5$

$\sf :\implies\:5a = -15$

$\sf :\implies\:a= \dfrac{-15}{5}$

$\bf :\implies\:a = -3ms^{-2}$

{Here -ve sign shows deceleration}

Hence, The acceleration of the bus before stopping is 3m/s².

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Now, Using Third Equation of Motion,

$\sf :\implies\:v^2-u^2 = 2as$

Here,

• v = 0m/s

• u = 15m/s

• a = -3m/s²

Putting Values,

$\sf :\implies\:0^2-15^2 = 2\times(-3)\times s$

$\sf :\implies\:225= -6 s$

$\sf :\implies\:s=\dfrac{225}{-6}$

$\bf :\implies\:s=-37.5\:m$

Hence, The distance travelled by bus before stopping is 37.5m.

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