Question

Find the acceleration of a body whose velocity increases from 11 m/s to 33 m/s in 10 seconds?

Answer 1

final velocity :33m/s , initial velocity :11m/s\

We know, a = v₂ - v₁ / t

Here, (v₂-v₁) = 33 - 11

= 22 m/s t = 10 s

Substitute their values,

a = 22/10 a = 2.2 m/s²

Hope this helps!  

Answer 2

Given :

•The velocity increases from 11 m/s to 33 m/s.

•Time = 10 seconds

To Find :

The acceleration of a body =?

Formula :

$\boxed{\sf{Acceleration = \dfrac{ \delta V}{\delta T} =\dfrac{V_{2}-V_{1} }{T_{2}-T_{1} } }}$

Solution :

$\implies \sf \: acceleration = \frac{ \delta \: v}{ \delta \: t} \\ \\ \\ \implies \sf \: acceleration = \frac{33 - 11}{10} \\ \\ \\ \implies \sf \: acceleration = \frac{22}{10} = 2.2 m {s}^{ - 2} \\ \\ \\ \implies \boxed{ \sf{acceleration = 2.2m {s}^{ - 2} }}$

The value of acceleration is 2.2m/s²