Question

A motor cyclist starts from rest and moves with uniform acceleration
of 1.9 m/t for 22 seconds
. Find the velocity and distance covered

Answer 1

Answer:

v = 41.8 m/s

s = 459.8 m

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 0 m/s (As it starts from rest)
• Acceleration (a) = 1.9 m/s²
• Time taken (t) = 22 seconds

We are asked to calculate final velocity (v) and distance covered (s).

Calculating final velocity (v) :

By using the first equation of motion,

⇒ v = u + at

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

⇒ v = 0 + 1.9(22)

⇒ v = 0 + 41.8

⇒ v = 41.8 m/s

∴ Final velocity of the motor cyclist is 41.8 m/s.

Calculating distance covered :

By using the second equation of motion :

⇒ s = ut + ½at²

• s denotes distance
• u denotes initial velocity
• t denotes time
• a denotes acceleration

⇒ s = 0(22) + ½ × 1.9 × (22)²

⇒ s = ½ × 1.9 × 484

⇒ s = 1 × 1.9 × 242

⇒ s = 459.8 m

∴ Distance covered by the motor cyclist is 459.8 m.

$\rule{200}2$

Extra Shots!!

Equations of motion :

$\boxed{ \begin{array}{cc} \pmb{\sf{ \quad \: v = u + at \quad}} \\ \\ \pmb{\sf{ \quad \: s= ut + \cfrac{1}{2}a{t}^{2} \quad \: } } \\ \\ \pmb{\sf{ \quad \: {v}^{2} - {u}^{2} = 2as \quad \:}}\end{array}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time
• s denotes distance