Physics, asked by yash5899, 4 months ago

4. A racing car has a uniform
acceleration of 4 m s?, What
distance will it cover in 10 s after
start?​

Answers

Answered by araziq522
1

Answer:

initial velocity zero

s=v+.5at^2

answer=200m

Answered by iTzShInNy
7

 \large \mathfrak { \underline{ \red {Correct \: Question \: }}}:- \\

 \small \rm A  \: racing  \: car  \: has \:  a  \: uniform  \: acceleration  \: of \\  \small \rm 4 \: ms {}^{ - 2} .What  \: distance \: will  \: it  \: cover \: in \: 10 \: s \:  \\  \small \rm after \: start  \: ? \\

 \\

 \large \mathfrak { \underline{ \red {Given \: }}}:- \\

  • Initial velocity , u ⇏ 0 m/sec

  • Acceleration , a ⇏ 4 m/ sec²

  • Time taken , t ⇏ 10 sec

 \\  \\

 \large \mathfrak { \underline{ \red {To\: Find}}}:- \\

  • Distance covered by the car , s ⇏ ?

 \\  \\

 \large \mathfrak { \underline{ \red {Formula \: }}}:- \\

 \small \star \bigstar \rm \: s⇢ ut +  \frac{1}{2}at {}^{2}   \bigstar \star \\

 \\  \\

 \large \mathfrak { \underline{ \red {Substituting \: }}}:- \\

We know,

  \small\rm \:↬ s⇢ ut +  \frac{1}{2}at {}^{2}   \\

 \small\rm \:↬ s⇢ 0 \times 10 +  \frac{1}{2} \times 4 \times  {10}^{  2}   \\

 \small\rm \:↬ s⇢ 0 +  \frac{1}{ \cancel2  } \times  \cancel4 \times  {100}  \\

 \small\rm \:↬ s⇢ 2 \times 100  \\

 \small\rm \:↬ s⇢ 200 \: m\\

Hence, the distance that the racing car will cover in 10 s after start is 200 m.

 \\  \\

 \large \mathfrak { \underline{ \red {Full \: Explaination \: }}}:- \\

  • In this query, Acceleration, Initial velocity and Time taken is given .

  • As the car starts from the rest so the Initial velocity will be 0 m/sec.

  • Here, it is given that we have to find out the Distance , (s) .

  • Using Newton's law of motion we can find the required answer.

  • As here distance (s) is given , we know two formulae to find it. That is :- 2nd Equation of motion and Third equation of motion.

  • As in the Third equation of motion , we need Final velocity ( v ) which is not mentioned in this query.

  • So , the right formula to use here is the 2nd equation of motion.

  • Applying the 2nd equation of motion , we will get the required Value.

 \\  \\

 \\

 \pink\bigstar{ \underline { \underline { \bf{↬ \purple{ ExpLore \: MoRe \: ForMuLa}↫}}}} \green \bigstar \\

 \small \green\bigstar {\underline {\boxed { \bf  {s = v \times t }}}}  \red\bigstar

where ,

  • s ➟ distance

  • v ➟ speed

  • t ➟ time taken

 \\

───※ ·❆· ※───

 \\

 \small \green\bigstar {\underline {\boxed { \bf  {v=u + at}}}}  \red\bigstar

where,

  • u ➟ Initial velocity

  • v ➟ Final velocity

  • a ➟ Acceleration

  • t ➟ time taken

 \\

───※ ·❆· ※───

 \\

 \small \green\bigstar {\underline {\boxed { \bf  {s=ut +  \frac{1}{2}at {}^{2}  }}}}  \red\bigstar

where,

  • s ➟ distance

  • u ➟ Initial velocity

  • t ➟ time taken

  • a ➟ Acceleration

 \\

───※ ·❆· ※───

 \\

 \small \green\bigstar {\underline {\boxed { \bf  {2as=v {}^{2}  -  {u}^{2}  }}}}  \red\bigstar \\

where,

  • a ➟ acceleration

  • s ➟ distance

  • v ➟ final velocity

  • u ➟ initial velocity

 \\

───※ ·❆· ※───

 \\

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