Question

A truck running at 90 km/h, slows down to 54 km/h over a distance of 20 m. Calculate (i) the retardation produced by its brakes, and (ii) the time for which the brakes are applied.

Answer 1

Answer:

Retardation = 10 m/s²

Time = 1 s

Explanation:

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

• Initial velocity (u) = 90 km/h
• Final velocity (v) = 54 km/h
• Distance covered (s) = 20 m

We are asked to calculate,

• The retardation of the truck.
• The time for which the brakes are applied.

Before commencing the steps, we need to convert velocities in its SI unit, that is m/s.

Converting initial velocity to m/s :

$\longmapsto \rm {u = 90 \; kmh^{-1} }$

• 1 km/h = 5/18 m/s

$\longmapsto \rm {u = \Bigg ( 90 \times \dfrac{5}{18} \Bigg ) \; ms^{-1} }$

$\longmapsto \rm {u = \Big ( 5 \times 5\Big ) \; ms^{-1} }$

$\longmapsto \bf {u = 25 \; ms^{-1} }$

Converting final velocity to m/s :

$\longmapsto \rm {v = 54 \; kmh^{-1} }$

• 1 km/h = 5/18 m/s

$\longmapsto \rm {v= \Bigg ( 54 \times \dfrac{5}{18} \Bigg ) \; ms^{-1} }$

$\longmapsto \rm {v = \Big ( 3 \times 5\Big ) \; ms^{-1} }$

$\longmapsto \bf {v= 15 \; ms^{-1} }$

$\rule{200}2$

Calculating retardation :

In order to calculate retardation, we need to find the acceleration first.

By using the third equation of motion :

$\longmapsto \bf{ v^2 - u^2 = 2as}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

$\longmapsto \rm { (15)^2 - (25)^2 = 2a(20)}$

$\longmapsto \rm { 225 - 625 = 40a}$

$\longmapsto \rm { -400 = 40a}$

$\longmapsto \rm { \cancel{\dfrac{-400}{40}} = a}$

$\longmapsto \rm { -10 \; ms^{-2} = a}$

Now, as the acceleration is –10 m/s². So, the retardation will be 10 m/s².

∴ Retardation is 10 m/s².

$\rule{200}2$

Calculating time taken :

Now, by using the first equation of motion :

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

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

$\longmapsto \rm { 15 = 25 + (-10)t}$

$\longmapsto \rm { 15 -25 = -10t}$

$\longmapsto \rm { -10= -10t}$

$\longmapsto \rm { \dfrac{-10}{-10} = t}$

$\longmapsto \bf { 1 \; s = t}$

∴ The time for which the brakes are applied is 1 second.

Answer 2

Answer:

10m/s thanks ❤️❤️