Question

A body starting from rest acquires a velocity of 10ms in 5 seconds.Calculate (a) the accelaration(b)the distance covered by the body in 5 seconds.

Answer 1

Given :

• Initial velocity (u) = 0 m/s (as it starts from rest)
• Final velocity (v) = 10 m/s
• Time (t) = 5 seconds

To find :

• Acceleration (a) &
• Distance (s)

According to the question,

At first we will find the acceleration using Newton's first equation of motion then we will find the distance using Newton's third equation of motion.

→ v = u + at

Where,

• v stands for Final velocity
• u stands for Initial velocity
• t stands for Time
• a stands for Acceleration

Substituting the values,

→ 10 = 0 + a(5)

→ 10 - 0 = 5a

→ 10 = 5a

→ 10 ÷ 5 = a

→ 2 = a

.°. The acceleration is 2 m/s²...

b)

By using Newton's third equation of motion,

→ s = ut + ½ at²

Where,

• s stands for Distance
• u stands for Initial velocity
• v stands for Final velocity
• t stands for Time

Substituting the values,

→ s = 0 × 5 + ½ × 2 × (5)²

→ s = 0 + 25

→ s = 25

.°. The distance is 25 m...

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