Help help help help please please e
1 answer:
The answer, when rounded to the nearest tenth is 19.2
First you can get rid of the decimals by multiplying both numbers by 100, this just makes the whole problem easier. Now all that needs to be done is long division:
(In my work I didn’t realize that it said round to the nearest tenth until I had already gotten past what I needed to find)
First you can get rid of the decimals by multiplying both numbers by 100, this just makes the whole problem easier. Now all that needs to be done is long division:
(In my work I didn’t realize that it said round to the nearest tenth until I had already gotten past what I needed to find)
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<em>Acceleration (a) = 5.1 m/s</em>²
<em>Initial speed (u) = 16 km/h </em>⇒
<em>m/s </em>≈ <em>4.5m/s</em><em>
</em><em>Final speed (v) =118 km/h </em>⇒
<em>m/s</em> ≈ <em>32.8m/s</em><em>
</em>
Distance(S) travel in that particular instant is carried out by 'Third equation of motion' i.e., v² = u² + 2aS
<em>So, When all quantities are in S.I. unit then,
</em>putting the values in the equation of motion,
<em /><em>As we have to carry out the distance covered,
</em><em>2</em>·<em>a</em>·<em>S = v</em>² <em>- u</em>²
<em>S = </em>
Putting values in derived equation,
⇒ <em>S = </em>
⇒ <em>S = </em>
⇒ <em>S = </em>
⇒ <em>S </em>≈ <em>103.489</em> <em>m
</em><em>
The total distance covered in that given condition is approx. 103.289 m.</em>
<em>Initial speed (u) = 16 km/h </em>⇒
</em><em>Final speed (v) =118 km/h </em>⇒
</em>
Distance(S) travel in that particular instant is carried out by 'Third equation of motion' i.e., v² = u² + 2aS
<em>So, When all quantities are in S.I. unit then,
</em>putting the values in the equation of motion,
<em /><em>As we have to carry out the distance covered,
</em><em>2</em>·<em>a</em>·<em>S = v</em>² <em>- u</em>²
<em>S = </em>
Putting values in derived equation,
⇒ <em>S = </em>
⇒ <em>S = </em>
⇒ <em>S = </em>
⇒ <em>S </em>≈ <em>103.489</em> <em>m
</em><em>
The total distance covered in that given condition is approx. 103.289 m.</em>