Question

After moving at a rate of 72 km per hour, con applies beaches which provide a recordation of 5 metre per Second Square How much times does the car take to Stop & How much distance does the car cover before. coming to rest?​

Answer 1

Appropriate Question:

After moving at a rate of 72 km/h, a car applies brakes which provide a retardation of 5 m/s². How much times does the car take to stop & How much distance does the car cover before coming to rest?

Explanation:

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

• Initial velocity (u) = 72 km/h
• Final velocity (v) = 0 m/s (As it stops)
• Acceleration (a) = –5 m/s² (As retardation is +5 m/s²)

We are asked to calculate time taken to stop and distance travelled before coming to rest.

Before commencing the steps, let's convert initial velocity in m/s.

$\longmapsto\rm { u = 72 \; kmh^{-1}}$

$\longmapsto\rm { u = \Bigg ( 72 \times \dfrac{5}{18} \Bigg ) \; ms^{-1}}$

$\longmapsto\rm { u = \Bigg ( 4 \times 5\Bigg ) \; ms^{-1}}$

$\longmapsto\bf { u = 20 \; ms^{-1}}$

★ Time taken to stop :

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

$\longmapsto\rm { 0= 20 + (-5t )}$

$\longmapsto\rm { 0 - 20= -5t }$

$\longmapsto\rm { -20 = -5t }$

$\longmapsto\rm { \dfrac{-20}{-5} = t }$

$\longmapsto\bf { 4 \; = t }$

∴ Time taken to stop is 4 seconds.

★ Distance travelled before coming to rest :

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

$\longmapsto\rm {(0)^2 - (20)^2 = 2(-5)s }$

$\longmapsto\rm { 0 - 400 = -10s }$

$\longmapsto\rm { - 400 = -10s }$

$\longmapsto\rm { \dfrac{ - 400 }{-10}= s }$

$\longmapsto\bf { 40 \; m= s }$

∴ Distance travelled before coming to rest is 40 m.