Question

The displacement of a particle is given by x = (t -2)2 where x is in meters and t is in seconds. The distance covered by the particle in first 4 seconds is (a)4m (b)8m (c) 12m (d) 16m

Answer 1

Given :

▪ The displacement of a particle is given by x = (t - 2)$^2$

• x is in meter
• t is in second

To Find :

▪ The distance covered by the particle in first 4 seconds.

Solution :

▶ Given : x = (t - 2)$^2$

At t = 0 : x = xo = (0-2)^2 = 4m

At t = 1s : x = x1 = (1-2)^2 = 1m

At t = 2s : x = x2 = (2-2)^2 = 0m

At t = 3s : x = x3 = (3-2)^2 = 1m

At t = 4s : x = x4 = (4-2)^2 = 4m

✏ The distance covered by the particle in 1st second is

D1 = xo - x1 = 3m

✏ The distance covered by the particle in 2nd second is

D2 = x1 - x2 = 1m

✏ The distance covered by the particle in 3rd second is

D3 = x3 - x2 = 1m

✏ The distance covered by the particle in 4th second is

D4 = x4 - x3 = 3m

✳ The distance covered by the particle in first 4 second is

D = D1 + D2 + D3 + D4

D = 3m + 1m + 1m + 3m

D = 8m

Answer 2

✪GiveN :-

The displacement of a particle is given by x = (t -2)² where,

✯ x is in meters, and

✯ t is in seconds.

✪TO FinD :-

➥ The distance covered by the particle in first 4 seconds.

✪AcknowledgemenT :-

It is given that x = (t -2)²

So, we can find the magnitude of displacements at different time intervals by putting the value of t to be from 1 to 4.

✪SolutioN :-

Putting the values of t :-

⇝ t = 0,

x₁ = (0 - 2)²

x₁ = (-2)²

x₁ = 4 m

$\rule{150}{2}$

⇝ t = 1,

x₂ = (1 - 2)²

x₂ = (-1)²

x₂ = 1 m

$\rule{150}{2}$

⇝ t = 2,

x₃ = (2 - 2)²

x₃ = (0)²

x₃ = 0 m

$\rule{150}{2}$

⇝ t = 3,

x₄ = (3 - 2)²

x₄ = (1)²

x₄ = 1 m

$\rule{150}{2}$

⇝ t = 4,

x₅ = (4 - 2)²

x₅ = 2²

x₅ = 4

$\rule{150}{2}$

Now,

➣ Distance covered in the 1st second = x₁ - x₂ = 4 - 1 = 3 m

➣ Distance covered in the 2nd second = x₂ - x₃ = 1 - 0 = 1 m

➣ Distance covered in the 3rd second = x₃ - x₄ = 0 - 1 = -1 m

➣ Distance covered in the 4th second = x₄ - x₅ = 1 - 4 = -3 m

In the 3rd and the 4th second, the sign of distance is negative. This justifies that the particle moved in the opposite direction. But we know, moving in the opposite direction affects the displacement not the distance. So, the negative sign will be avoided here.

$\rule{150}{2}$

Now, the distance covered by the particle in first 4 seconds is = 3 + 1 + 1 + 3

= 8 m

So, the correct answer is (b) 8 m.