Question

A particle starts from rest and has an acceleration of 2m per s^2 for 10 sec . After that, it travels for 30 seconds with constant speed and then under goes a retardation of 4m per s^2 and comes back to rest. find the distance travelled by the particle .

Answer 1

Ans: 750 m ......
....
...
.

Answer 2

Answer:

Explanation:

Solution,

Here, we have

For motion with acceleration, u = 0 m/s

Acceleration, a = 2 m/s²

Time taken, t = 10 seconds.

To Find,

Total distance covered, s = ?

Here, we know that

s₁ = ut + 1/2at²

⇒ s₁ = 0 + 1/2 × 2 × (10)²

⇒ s₁ = 100 m.

For motion with constant speed,

t = 30 s

s₂ = vt

⇒ s₂ = 20 × 30

⇒ s₂ = 600 m.

For the motion with retardation,

u = 20 m/s

a = - 4 m/s²

v = 0 m/s

As we know that

v² - u² = 2as

⇒ (0)² - (20)² = 2 × (- 4) × s₃

⇒ 400 = - 8s₃

⇒ s₃ = 50 m.

Now, the total distance covered,

⇒ s = s₁ + s₂ + s₃

⇒ s = (100 + 600 + 50) m

⇒ s = 750 m

Hence, the distance covered by the particle is 750 m.