Question

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

Answer 1

Answer:

Explanation:

Solution,

Here, we have

Initial velocity, u = 6 m/s

Final velocity, v = 0 m/s

Acceleration, a = - 1 m/s²

To Find,

a. Distance covered, s = ?

b. Time taken, t = ?

According to the 3rd equation of motion,

v² - u² = 2as

So, putting all the values, we get

⇒ v² - u² = 2as

⇒ (0)² - (6)² = 2 × (- 1) × s

⇒ 36 = - 2s

⇒ 36/2 = s

⇒ s = 18 m

Hence, the distance covered by the object is 18 m.

According to the 1st equation of motion,

We know that,

v = u + at

So, putting all the values again, we get

⇒ v = u + at

⇒ 0 = 6 + (- 1) × t

⇒ t = 6 seconds.

Hence, the time taken to coming to rest is 6 seconds.