Question

A body starts from rest from the origin with an acceleration of 3 m//s(2) along the x-axis and 4 m//s^(2) along the y-axis. Its distance from the origin after 2 s will be

Answer 1

$\huge\fcolorbox{red}{pink}{Answer}$

✏ Given:

$\rm \: initial \: velocity = 0 \: \frac{m}{s} \\ \rm \: acceleration \: along \: x \: axis = 3 \hat{i} \: \frac{m}{ {s}^{2} } \\ \rm \: acceleration \: along \: y \: axis = 4 \hat{j} \: \frac{m}{ {s}^{2} }$

✏ To Find:

Distance from origin after 2 s

✏ Formula:

Distance covered by body in acceleratory motion is given by

$\: \: \: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \bold{ \rm{ \pink{d = ut + \frac{1}{2} a {t}^{2} }}}}} \: \dag$

✏ Calculation:

$\implies \rm \: d = (0 \times 2) + \frac{1}{2} \times (3 \hat{i} + 4 \hat{j}) \times ( {2})^{2} \\ \\ \therefore \rm \: d = 2(3 \hat{i} + 4 \hat{j}) \\ \\ \therefore \: \underline{ \boxed{ \red{ \rm{d = 6 \hat{i} + 8 \hat{j} = 10 \: m}}}}$

Answer 2

Answer:

• The distance (d) after 2 sec will be 10 meters.

Given:

1. Acceleration along x-axis (a_x) = 3 m/s²
2. Acceleration along y-axis (a_y) = 4 m/s²
3. Initial velocity (u) = 0 m/s.

Explanation:

$\rule{300}{1.5}$

From the relation we know,

⇒ a = √ (a_x)² + (a_y)²

Where,

• a Denotes Acceleration.
• a_x Denotes acceleration at x-axis.
• a_y Denotes Acceleration at y-axis.

Now,

⇒ a = √ (a_x)² + (a_y)²

Substituting the values,

⇒ a = √ (3)² + (4)²

⇒ a = √ 9 + 16

⇒ a = √ 25

⇒ a = 5

⇒ a = 5 m/s².

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Second Kinematic Equation we know,

⇒ S = u t + 1 / 2 a t²

Where,

• S Denotes Displacement.
• u Denotes Initial velocity.
• t Denotes Time taken.
• a Denotes Acceleration.

Now,

⇒ S = u t + 1 / 2 a t²

Substituting the values,

⇒ d = 0 × 2 + 1 / 2 × 5 × (2)²

∵ [ S = d ; t = 2 sec ; a = 5 m/s² ]

⇒ d = 0 + 1 / 2 × 5 × 4

⇒ d = 1 / 2 × 5 × 4

⇒ d = 5 × 2

⇒ d = 10

⇒ d = 10 m.

∴ The distance (d) after 2 sec will be 10 meters.

$\rule{300}{1.5}$