Question

when brakes are applied to a bus, the retardationproduces is 25cm/s² and the bus takes 20 sec to stop . claculate the initiak velocity of tue bus and the distance travelled in 5seconds.
plz help!​

Answer 1

Answer:

Initial velocity = 5 m/s

Distance travelled in 5 seconds = 21.875 m

Explanation:

As per the provided information in the question, we have :

• Retardation = 25 cm/s²
• Time taken (t) = 20 seconds
• Final velocity (v) = 0 (as it stops)

Here, as retardation is 25 cm/s², so the acceleration will be –25 cm/s².

$\longmapsto \rm { a = -25 \; cm \: s^{-2} }$

• 1 cm/s² = 0.01 m/s²

$\longmapsto \rm { a = (-25 \times 0.01) \; m \: s^{-2} }$

$\longmapsto \bf { a =- 0.25 \; m \: s^{-2} }$

$\rule{200}2$

Calculating initial velocity :

By using the first equation of motion,

$\longmapsto \bf {v = u + at }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\longmapsto \rm {0 = u + \Big( -0.25 \times 20 \Big ) }$

$\longmapsto \rm {0 = u + \Bigg( -\dfrac{25}{100} \times 20 \Bigg ) }$

$\longmapsto \rm {0 = u + \Bigg( -\dfrac{25}{10} \times 2 \Bigg ) }$

$\longmapsto \rm {0 = u + \Bigg( -\dfrac{25}{5} \Bigg ) }$

$\longmapsto \rm {0 = u + (-5 ) }$

$\longmapsto \rm {0 = u -5}$

$\longmapsto \rm {0 +5= u}$

$\longmapsto \bf {5 \; ms^{-1}= u}$

∴ Initial velocity of the bus is 5 m/s.

$\rule{200}2$

By using the second equation of motion,

$\longmapsto \bf { s = ut + \dfrac{1}{2}at^2}$

• s denotes distance
• u denotes initial velocity
• t denotes time
• a denotes acceleration

Here, time will be 5s as we have to find the distance travelled in 5seconds.

$\longmapsto \rm { s = 5(5) + \Bigg \{ \dfrac{1}{2} \times (-0.25) \times (5)^2 \Bigg \} }$

$\longmapsto \rm { s = 25 + \Bigg \{ \dfrac{1}{2} \times \Bigg( -\dfrac{25}{100} \Bigg) \times 25 \Bigg \} }$

$\longmapsto \rm { s = 25 + \Bigg \{ \dfrac{1}{2} \times \Bigg( -\dfrac{1}{4} \Bigg) \times 25 \Bigg \} }$

$\longmapsto \rm { s = 25 + \Bigg \{ -\dfrac{1}{8} \times 25 \Bigg \} }$

$\longmapsto \rm { s = 25 + \Bigg \{ -\dfrac{25}{8} \Bigg \} }$

$\longmapsto \rm { s = 25 -\dfrac{25}{8} }$

$\longmapsto \rm { s = \dfrac{200 - 25}{8} }$

$\longmapsto \rm { s = \dfrac{175}{8} }$

$\longmapsto \bf { s = 21.875 \; m}$

∴ The distance travelled in 5 seconds is 21.875.

$\rule{200}2$