Question

An object is moving with a velocity of + 6 m s-^1 and
with an acceleration of -1 m s-^2.What will be the
(a) distance travelled by the car before coming to
rest?
(b) time taken for coming to rest?​

Answer 1

Given Information :

• Initial velocity (u) = 6 m/s

• Acceleration (a) = – 1 m/s²

To calculate :

• Distance travelled by the car before coming to rest.

• Time taken for coming to rest.

Calculation :

★ Distance travelled by the car before coming to coming to rest :

By using third equation of motion,

• v² = u² + 2as

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

Since, car will come to rest. So, its final velocity can be considered as 0 m/s.

→ (0)² = (6)² + {2 × (-1) × s}

→ 0 = 36 + (-2)s

→ 0 = 36 - 2s

→ 0 + 2s = 36

→ 2s = 36

→ s = $\sf { \cancel { \dfrac{36}{2} }}$

➝ s = 18 m

Therefore, distance travelled by the car 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 }

→ 0 = 6 - t

→ 0 + t = 6

➝ t = 6 s

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

More Information :

Some more formulae :

First equation of motion :

• v = u + at

Second equation of motion :

• s = ut + ½at²

Third equation of motion :

• v² = u² + 2as

Where,

• v denotes final velocity

• u denotes initial velocity

• a denotes acceleration

• s denotes distance

• t denotes time