Question

(13) A body moving with a velocity of 50 cm s- undergoes a uniform
acceleration of 20 cms-2. How far does it move in 4 s and what is its velocity
after that time?​

Answer 1

Given :-

▪ A body is moving with a velocity of u = 50 cm/s and it undergoes a uniform acceleration of a = 20 cm/s²

To Find :-

▪ How far does it move after 4 seconds.

▪ Velocity of the body after 4 seconds.

Solution :-

We are given,

• Initial velocity, u = 50 cm/s
• Acceleration, a = 20 cm/s
• Time, t = 4 s

Using second equation of motion,

⇒ s = ut + 1/2 at²

⇒ s = 50 × 4 + 1/2 × 20 × (4)²

⇒ s = 200 + 10×16

⇒ s = 200 + 160

⇒ s = 360 cm

Now, Let us find the velocity after 4 seconds,

Using first equation of motion, we have

⇒ v = u + at

⇒ v = 50 + 20×4

⇒ v = 50 + 80

⇒ v = 130 cm/s

Hence, The body would go 360 cm from its initial position and the velocity of the body will be 130 cm/s

Some Formulae :-

☞ Equations of motion,

• 1st : v = u + at
• 2nd : s = ut + 1/2 at²
• 3rd : 2as = v² - u²

☞ Free fall body equations,

• Time of flight = 2u / g
• Max height = u² / 2g

Answer 2

Given:

• Initial velocity=u = 50cm/s
• Acceleration= a = 20 $cm/{s}^{2}$
• Time = t= 4s

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Need to find :

• Final velocity = v=?
• Distance travelled = s=?

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Solution :

From 1st equation of motion we know,

$v = u + at$

$\implies \: v = 50 + 20 \times 4 cm/s$

$\implies \: v = 50 + 80 cm/s$

$\red{\bf{ \implies \: v = 130cm/s}}$

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From 2nd equation of motion we know,

$s = ut + \dfrac{1}{2}a {t}^{2} cm$

$\implies \: s = 50 \times 4 + \dfrac{1}{2} \times 20 \times {4}^{2} cm$

$\implies \: 200 + 160cm$

$\red{\bf{ \implies \: s = 360cm}}$