Question

A particle starts to move in a straight line from a point with velocity:
$10ms {}^{ - 1}$
and acceleration:
$- 2.0ms {}^{ - 2}$
Find the position and velocity of the particle at
(i) t = 5s
(ii) t = 10 s
________________________

✔️ Explained answers only
❌ No copied answer


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Answer 1

Answer:

1 option is true I know I saw that question so many times in my book

Explanation:

ok a

Answer 2

$\huge{ \underbrace{ \text{Question}}}$

A particle starts to move in a straight line from a point with velocity:

$\sf{10ms {}^{ - 1} }$

and acceleration:

$\sf{ - 2.0ms {}^{ - 2} }$

Find the position and velocity of the particle at

(i) t = 5s

(ii) t = 10 s

$\huge{ \underbrace{ \text{Answer}}}$

Given:-

$\sf{ \rightarrow{Initial \: velocity \: \bold{u}\: of \: the \: article = 10ms {}^{ - 1} }}$

$\sf{ \rightarrow{Acceleration \: \bold{a} \: of \: the \: article \: = - 2.0ms {}^{ - 2} }}$

To Find:-

$\sf{ \rightarrow{The \: position \: and \: velocity \: of \: the \: article \: when, }}$

$\sf{i)t = 5s} \\ \sf{ii)t = 10s}$

Solution:-

First, let's find the velocity of the article.

As we know that,

$\boxed{ \sf{ \blue{ \bold{v} = u + at}}}$

Where,

• v stands for velocity
• u stands for initial velocity
• a stands for acceleration
• t stands for time

Now, according to the question,

If t = 5s,

$\sf{ \bold{v} = 10ms {}^{ - 1} + ( - 2.0 {ms}^{ - 2}) \times 5s}$

$\\ \sf{ \implies{ \bold{v} = 10ms {}^{ - 1} - 2.0 {ms}^{ - 2} \times 5s}}$

$\\ \sf{ \implies{\bold{v} = 10ms {}^{ - 1} - 10ms {}^{ - 1} }}$

$\\ \sf{ \therefore{ \bold{ v} = \orange{ 0ms {}^{ - 1} }}}$

Again,

If t = 10s,

$\sf{ \bold{v} = 10ms {}^{ - 1} + ( - 2.0 {ms}^{ - 2}) \times 10s}$

$\\ \sf{ \implies{ \bold{v} = 10ms {}^{ - 1} - 2.0 {ms}^{ - 2} \times 10s}}$

$\\ \sf{ \implies{\bold{v} = 10ms {}^{ - 1} - 20ms {}^{ - 1} }}$

$\\ \sf{ \therefore{ \bold{ v} = \orange{ - 10ms {}^{ - 1} }}}$

Now,

Let's find the position of the article.

As we know that,

$\sf{ \boxed{ \blue{ \bold{s = ut + \frac{1}{2} a {t}^{2} }}}}$

Where,

• s stands for position
• u stands for initial velocity
• a stands for acceleration
• t stands for time

Now, according to the question,

If t = 5s

$\sf{ \bold{s }= 10 {ms}^{ - 1} \times 5s + \frac{1}{2} \times ( - 2.0 {ms}^{ - 2} ) \times {5s}^{2} }$

$\\ \sf{ \implies{ \bold{s }= 50m + \frac{1}{2} \times ( - 50m)}}$

$\\ \sf{ \implies{ \bold{s} = 50m +( - 25m)}}$

$\\ \sf{ \implies{ \bold{s} = 50m - 25m}}$

$\\ \sf{ \therefore{ \bold{s} = \orange{25m}}}$

Again,

If t = 10s,

$\sf{ \bold{s }= 10 {ms}^{ - 1} \times 10s + \frac{1}{2} \times ( - 2.0 {ms}^{ - 2} ) \times {10s}^{2} }$

$\\ \sf{ \implies{ \bold{s }= 100m + \frac{1}{2} \times ( - 200m)}}$

$\\ \sf{ \implies{ \bold{s} = 100m +( - 100m)}}$

$\\ \sf{ \implies{ \bold{s} = 100m - 100m}}$

$\\ \sf{ \therefore{ \bold{s} = \orange{0m}}}$

$\underline{ \rm{Hence \: velocity \: of \: the \: article \: is \: \bold{ 0 {ms}^{ - 1}} \: when \: \bold{t = 5s}}}$

$\underline{ \rm{Velocity \: of \: the \: article \: is \: \bold{ - 10{ms}^{ - 1}} \: when \: \bold{t = 10s}}}$

$\underline{ \rm{Position \: of \: the \: article \: is \: \bold{ 25m} \: when \: \bold{t = 5s}}}$

$\underline{ \rm{Position \: of \: the \: article \: is \: \bold{ 0m} \: when \: \bold{t = 10s}}}$