Question

A Car is moving with a Velocity of 6 m/s and with a acceleration of -1.m's square what will be the distance travelled by car before coming to rest? Time Taken for coming to rest ? ​

Answer 1

Answer:

Distance travelled by car before coming to rest = 18 m

Time Taken for coming to rest = 6s

Explanation:

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

• Initial velocity (u) = 6 m/s
• Acceleration (a) = -1 m/s²
• Final velocity (v) = 0 m/s (As it comes to rest)

We are asked to calculate the distance travelled by car before coming to rest and time taken for coming to rest.

The distance travelled by car before coming to rest :

By using the third equation of motion,

→ v² - u² = 2as

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

→ (0)² - (6)² = 2 × (-1) × s

→ 0 - 36 = -2s

→ -36 = -2s

→ -36 ÷ (-2) = s

→ 18 m = s

Therefore, distance travelled before coming to rest is 18 m.

Time Taken for coming to rest :

By using the first equation of motion,

→ v = u + at

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

→ 0 = 6 + (-1)t

→ 0 - 6 = -1t

→ -6 = -1t

→ -6 ÷ (-1) = t

→ 6 s = t

Therefore, time taken for coming to rest is 6 seconds .

$\rule{200}2$

More Information :

First equation of motion :

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

Second equation of motion :

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

Third equation of motion :

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

Where,

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