Question

A body is moving vertically upwards.Its velocity changes at a constant rate from 60m/s to 30m/s in 3 seconds. what is its acceleration?​

Answer 1

Answer:

$\bigstar$ Given:

⇒ Initial velocity ($\rm v_i$) = 60 m/s

⇒ Final velocity ($\rm v_f$) = 30 m/s

⇒ Velocity changes in time (Δt) = 3 sec

$\bigstar$ To Find:

⇒ Acceleration (a)

$\bigstar$ Solution:

To find the average Acceleration (a)

We use the general formula,

$\green{\boxed{\rm a = \dfrac{\Delta v}{\Delta t}}}$

Δv is the change in velocity ($\rm v_f - v_i$)

Therefore,

$\blue{\boxed{\rm a = \dfrac{v_f - v_i}{\Delta t}}}$

Substituting the above values in the formula

We get,

$\implies \rm a = \dfrac{30 - 60}{3} \;m/s^2$

$\implies \rm a = \dfrac{-30}{3} \;m/s^2$

$\implies \rm a = -10 \;m/s^2$

$\purple{\boxed{\rm {a = -10\;m/s^2}}}$

Therefore, the Acceleration

(a) = -10 m/s²

Answer 2

Given that, initial velocity (u) of the body is 60 m/s and final velocity (v) of the body is 30 m/s.

Also, given that time (t) is 3 seconds.

We have to find the acceleration (a) of the body.

Using First Equation of Motion,

⇒ v = u + at

Substitute the known values

⇛ 30 = 60 + a(3)

⇛ 30 - 60 = 3a

⇛ -30 = 3a

Divide by 3 on both sides

⇛ -30/3 = 3a/3

⇛ -10 = a

⇛ a = -10 m/s²

(Negative sign shows retardation pr deceleration)

$\rule{200}2$

Additional Information

Acceleration is defined as a change in an object's velocity over time.

i.e. a = ∆v/t or a = (v - u)/t

It's S.I. unit is m/s².