Question

II. A body is moving vertically upwards. Its
velocity changes at a constant rate from 50 m - to
20 ms- in 3 What is its acceleration ?​

Answer 1

$\underline{ \underline{ \Large \pmb{\mathit{ {Appropriate \: Question:}} }} }$

A body is moving vertically upwards. Its

velocity changes at a constant rate from $\sf {50 \: m \: {s}^{-1} }$ to $\sf {20 \: m \: {s}^{-1} }$ in 3 seconds. What is its acceleration ?

$\underline{ \underline{ \Large \pmb{\mathit{ {Given:}} }} }$

• Initial velocity (u) = $\sf {50 \: m \: {s}^{-1} }$

• Final velocity (v) = $\sf {20 \: m \: {s}^{-1} }$

• Time (t) = 3 seconds

$\underline{ \underline{ \Large \pmb{\mathit{ {To \: Find :}} }} }$

• Acceleration (a)

$\underline{ \underline{ \Large \pmb{\mathit{ {Calculation:}} }} }$

Method I ::

By using the formula of acceleration,

→ a = $\sf { \dfrac{v-u}{t} }$ m/s²

→ a = $\sf { \dfrac{20 - 50}{3} }$ m/s²

→ a = $\sf { \dfrac{-30}{3} }$ m/s²

→ a = - 10 m/s²

Therefore, acceleration of the body is - 10 m/s².

Method II ::

By using the first equation of motion,

→ v = u + at

→ 20 = 50 + 3a

→ 20 - 50 = 3a

→ - 30 = 3a

→ → a = $\sf { \dfrac{-30}{3} }$ m/s²

→ a = - 10 m/s²

Therefore, acceleration of the body is - 10 m/s².

$\underline{ \underline{ \Large \pmb{\mathit{ {Know \: More :}} }} }$

Equations of motion:

• v = u + at

• s = ut + ½at²

• v²– u² = 2as

Where,

• ★ v = final velocity = m/s
• ★ u = initial velocity = m/s
• ★ a = acceleration = m/s²
• ★ s = distance/displacement = m
• ★ t = time = sec

Remember that!

• When a body starts from rest, its initial velocity is 0.

• When a body comes to stop or applies breaks, its final velocity is 0.