Question

. A particle moves rectilinearly with a constant acceleration 1 m/s. Its speed after
10 seconds is 5 m/s. The distance covered by the particle in this duration is
1. 20m
2. 25m
3. 30m
4. 50m​

Answer 1

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

GiveN:

• Accleration of the particle = 1 m/s²
• Time taken = 10 s
• Final velocity = 5 m/s

To FinD:

• The distance covered by the particle in this duration is?

Step-wise-Step Explanation:

The above question can be solved by using the equations of motion:

1. v = u + at
2. s = ut + 1/2 at²
3. v² = u² + 2as

Finding initial velocity using 1st eq. of motion,

➛ v = u + at

➛ 5 = u + 1 × 10

➛ u = -5 m/s

Finding Distance using 2nd eq. of motion,

➛ v² = u² + 2as

➛ 5² = (-5)² + 2(1)s

➛ 25 + 25 = 2s

➛ 2s = 50 m

➛ s = 25 m

Hence,

The distance covered by the particle is:

$\huge{ \boxed{ \sf{ \red{25 \: m}}}}$

Answer 2

Answer:

Given :-

• A particle moves rectilinearly with a constant acceleration is 1 m/s and its speed after 10 seconds is 5 m/s.

To Find :-

• What is the distance covered by the particle.

Formula Used :-

❶ First equation no motion,

$\boxed{\bold{\large{v\: =\: u\: +\: at}}}$

❷ Second equation of motion,

$\boxed{\bold{\large{{v}^{2}\: =\: {u}^{2}\: +\: 2as}}}$

Solution :-

Given :

• Acceleration (a) = 1 m/s
• Time (t) = 10 secs
• Final velocity (v) = 5 m/s

❶ First we have to find the initial velocity (u),

✦ v = u + at

⇒ 5 = u + 1(10)

⇒ 5 = u + 10

⇒ 5 - 10 = u

⇒ - 5 = u

⇒ u = - 5 m/s

❷ Now, we have to find the distance ,

✦ v² = u² + 2as

⇒ (5)² = (- 5)² + 2(1) × s

⇒ 25 = 25 + 2 × s

⇒ 25 + 25 = 2s

⇒ 50 = 2s

⇒ - 2s = - 50

⇒ s = $\sf\dfrac{\cancel{- 50}}{\cancel{- 2}}$

➠ s = 25 m

$\therefore$ The distance covered by particles is 25 m .