Question

if the initial velocity of an object is 3 m/s and it changes to 13
m/s in 5 seconds then what is the value of acceleration? also find
the average velocity over the time period

Answer 1

• Initial velocity=u=3m/s
• Final velocity=v=13m/s
• Time=t=5s

We know

$\boxed{\sf Acceleration=\dfrac{v-u}{t}}$

$\\ \sf\longmapsto Acceleration= \dfrac{13-3}{5}$

$\\ \sf\longmapsto Acceleration=\dfrac{10}{5}$

$\\ \sf\longmapsto Acceleration=2m/s^2$

Answer 2

Answer:

Given,

$\sf \: \green \odot \: initial \: velocity \: \: u = 3 \: m {s}^{ - 1} \\ \sf \: \green \odot \: final \: velocity \: \: v = 13 \: m {s}^{ - 1} \\ \sf \: \green \odot \: time \: \: t = 5 \: s$

To find :

$\sf \: \green \odot \: acceleration \: \: = \: a \\ \: \\ \sf \: \green \odot \: average \: velocity \: = \: \bar {v}$

Formula that will be used :

$\sf \: \green \odot \: \sf \: acceleration \: \: \: a = \frac{v - u}{t} \\ \\ \sf \: \green \odot \: averge \: velocity \: \: \bar{v} = \frac{s}{t} \\ \\ \sf \: \green \odot \: distance \: \: s = ut + \frac{1}{2} a {t}^{2}$

Solution :

$\sf \: acceleration \: \: \: a \: = \frac{13 - 3}{ 5} \\ \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{10}{5} \\ \: \: \: \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 \: m {s}^{ - 2} \\ \\ \green{ \boxed{\bf \: so \: acceleration \: \: a = 2 \: m {s}^{ - 2} }}$

And

$\sf \: s \: = ut + \frac{1}{2} a {t}^{2} \\ \: \: \: \: \: = 3 \times 5 + \frac{1}{2} \times 2 \times {5}^{2} \\ \: \: \: \: \: \: = 40 \: \\ \\ \sf \: average \: velocity \: \: \: \bar{v} \: = \frac{s}{t} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{40}{5} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 8 \: m {s}^{ - 1} \\ \\ \green{ \boxed{\bf average \: velocity \: \: \bar{v} = 8 \: m {s}^{ - 1} }}$